Q: Question 1(vii)A car starting from rest moves on a straight path for time t = 0 to t = T with a uniform acceleration a and then stops with a uniform retardation. The average speed of the car will be :aT4\dfrac{\text {aT}}{4}4aT3aT2\dfrac{3\text {aT}}{2}23aTaT2\dfrac{\text {aT}}{2}2aTaT

aT2\dfrac{\text {aT}}{2}2aTReasonAssuming, uniform retardation has the same value as the uniform acceleration (a).Case 1 : When car is acceleratingGiven,Initial speed (u1) = 0Acceleration (a1) = aTime (t1) = TLet, distance travelled be S1 and final velocity is v1.As,v = u + atOn putting valuesv1 = u1 + a1t1 = 0 + aT = aTBy using,S = ut + 12\dfrac{1}{2}21at2On putting valuesS1 = u1t1 + 12\dfrac{1}{2}21a1t12\text t_1^2t12 = 0 + 12\dfrac{1}{2}21aT2 = 12\dfrac{1}{2}21aT2Case 2 : When car is retardingGiven,Initial speed (u2) = v1 = aTAcceleration (a2) = -aFinal speed (v2) = 0Let, distance travelled be S2 and time taken is t2.As,v = u + atOn putting valuesv2 = u2 + a2t2⟹ 0 = aT - at2⟹ aT = at2⟹ t2 = TBy using,S = ut + 12\dfrac{1}{2}21at2On putting valuesS2 = u2t2 + 12\dfrac{1}{2}21a2t22\text t_2^2t22= (aT)(T) - 12\dfrac{1}{2}21aT2= aT2 - 12\dfrac{1}{2}21aT2= 12\dfrac{1}{2}21aT2Here,Total distance travelled (S) = S1 + S2 = 12\dfrac{1}{2}21aT2 + 12\dfrac{1}{2}21aT2 = aT2Total time taken (t) = t1 + t2 = T + T = 2TThen,Avg. Speed (v)=Total distance (S)Total time (t)=aT22T=aT2\text {Avg. Speed (v)}=\dfrac{\text {Total distance (S)}}{\text {Total time (t)}} \\[1em] =\dfrac{\text {aT}^2}{2\text T}= \dfrac{\text {aT}}{2}Avg. Speed (v)=Total time (t)Total distance (S)=2TaT2=2aT∴ Average speed of the car is aT2.\dfrac{ {\bold {aT}}}{\bold 2}.2aT.

Q: Question 2A body starts with an initial velocity of 10 m s-1 and acceleration 5 m s-2. Find the distance covered by it in 5 s .

According to equation of motion —S = ut + 12\dfrac{1}{2}21 at2Where,distance covered = Sinitial velocity u = 10 m s-1acceleration a = 5 m s-2time t = 5 sSubstituting the values in the formula, we get,S=(10×5)+(12×5×52)S=(50)+(12×5×25)S=(50)+(2.5×25)S=50+62.5⇒S=112.5 m\text S = (10 \times 5) + (\dfrac{1}{2} \times 5 \times 5^2) \\[0.5em] \text S = (50) + (\dfrac{1}{2} \times 5 \times 25) \\[0.5em] \text S = (50) + ( 2.5 \times 25) \\[0.5em] \text S = 50 + 62.5 \\[0.5em] \Rightarrow \text S = 112.5\ \text m \\[0.5em]S=(10×5)+(21×5×52)S=(50)+(21×5×25)S=(50)+(2.5×25)S=50+62.5⇒S=112.5 mHence, distance covered = 112.5 m.

Q: Question 11A body moves from rest with a uniform acceleration and travels 270 m in 3 s. Find the velocity of the body at 10 s after the start.

As we know,S = ut + 12\dfrac{1}{2}21 at2Given,u = 0 m s -1S = 270 mt = 3 sSubstituting the values in the formula above we get,270=(0×3)+(12×a×32)270=92×a270×2=9×a⇒a=5409⇒a=60 ms−2270 = (0 \times 3) + (\dfrac{1}{2} \times \text a \times 3^2) \\[0.5em] 270 = \dfrac{9}{2} \times \text a \\[0.5em] 270 \times 2 = 9 \times \text a \\[0.5em] \Rightarrow \text a = \dfrac{540}{9} \\[0.5em] \Rightarrow \text a = 60\ \text{ms}^{-2}270=(0×3)+(21×a×32)270=29×a270×2=9×a⇒a=9540⇒a=60 ms−2Hence, a = 60 m s-2To find out velocity after t = 10 s.v = u + atSubstituting the values in the formula above we get,v=0+(60×10)⇒v=600 ms−1\text v = 0 + (60 \times 10) \\[0.5em] \Rightarrow \text v = 600\ \text{ms}^{-1} \\[0.5em]v=0+(60×10)⇒v=600 ms−1Hence, velocity after 10 s = 600 m s-1

Question 1

Question 15A car travels the first 25\dfrac {2}{5}52​th part of its total distance with a speed v1, and the remaining 35\dfrac {3}{5}53​th of the distance with speed v2. Its average speed is :12v1v2\dfrac{1}{2}\sqrt{	ext v_1 	ext v_2}21​v1​v2​​v1+v22\dfrac{	ext v_1 +	ext v_2}{2}2v1​+v2​​2v1v2v1+v2\dfrac{2	ext v_1 	ext v_2}{	ext v_1 +	ext v_2}v1​+v2​2v1​v2​​5v1v23v1+2v2\dfrac{5	ext v_1 	ext v_2}{3	ext v_1 +2	ext v_2}3v1​+2v2​5v1​v2​​

Accepted Answer

5v1v23v1+2v2\dfrac{5\text v_1 \text v_2}{3\text v_1 +2\text v_2}3v1​+2v2​5v1​v2​​Reason — Let, total distance covered be S of which S1 is covered with v1 in time t1 and S2 with v2 in time t2.Then,S1=25S\text S_1=\dfrac{2}{5}\text SS1​=52​S ............... (1)andS2=35S\text S_2=\dfrac{3}{5}\text SS2​=53​S ............... (2)AsSpeed (v)=Distance (S)Time (t)\text {Speed (v)}=\dfrac{\text {Distance (S)}}{\text {Time (t)}}Speed (v)=Time (t)Distance (S)​Then,⇒v1=S1t1⇒t1=S1v1\Rightarrow \text v_1=\dfrac{\text S_1}{\text t_1} \[1 em] \Rightarrow \text t_1=\dfrac{\text S_1}{\text v_1}⇒v1​=t1​S1​​⇒t1​=v1​S1​​Putting value of S1 from equation 1⇒t1=25Sv1\Rightarrow \text t_1 = \dfrac{2}{5}\dfrac{\text S}{\text v_1}⇒t1​=52​v1​S​Similarly⇒v2=S2t2[1em]⇒t2=S2v2\Rightarrow \text v_2=\dfrac{\text S_2}{\text t_2} \ [1 em] \Rightarrow\text t_2=\dfrac{\text S_2}{\text v_2}⇒v2​=t2​S2​​[1em]⇒t2​=v2​S2​​Putting value of S2 from equation 2⇒t2=35Sv2\Rightarrow\text t_2= \dfrac{3}{5}\dfrac{\text S}{\text v_2}⇒t2​=53​v2​S​ Now,Average Speed (v)=Total distance travelled (S)Total time taken (t)=St1+t2=S2S5v1+3S5v2=SS5(2v1+3v2)=SS5(2v2+3v1v1v2)=5v1v23v1+2v2\text {Average Speed (v)}=\dfrac{\text {Total distance travelled (S)}}{\text {Total time taken (t)}} \[1 em] =\dfrac{\text S}{\text t_1 + \text t_2} \[1 em] =\dfrac{\text S}{\dfrac{2\text S}{5\text v_1}+\dfrac{3\text S}{5\text v_2}} \[1 em] = \dfrac{\text S}{\dfrac{\text S}{5}\Big(\dfrac{2}{\text v_1} + \dfrac{3}{\text v_2}\Big)} \[1em] = \dfrac{\text S}{\dfrac{\text S}{5}\Big(\dfrac{2\text v_2 + 3 \text v_1}{\text v_1 \text v_2} \Big)} \[1em] =\dfrac{5\text v_1 \text v_2}{3\text v_1 +2\text v_2}Average Speed (v)=Total time taken (t)Total distance travelled (S)​=t1​+t2​S​=5v1​2S​+5v2​3S​S​=5S​(v1​2​+v2​3​)S​=5S​(v1​v2​2v2​+3v1​​)S​=3v1​+2v2​5v1​v2​​

Question 2

Question 6A car travels first 30 km with a uniform speed of 60 km h-1 and then next 30 km with a uniform speed of 40 km h -1.Calculate —(i) the total time of journey,(ii) the average speed of the car.

Accepted Answer

Given,Distance travelled S1 = 30 kmSpeed v1 = 60 km h-1Distance travelled S2 = 30 kmSpeed v2 = 40 km h-1(i) As we know,Time (t)=distance (S)speed (v)\text {Time (t)} = \dfrac {\text {distance (S)}}{\text {speed (v)}}Time (t)=speed (v)distance (S)​Substituting the values in the formula above, we get,t1=S1v1t1=30km60km h−1⇒t1=12h⇒t1=0.5h\text{t}_1 = \dfrac{\text{S}_1}{\text{v}_1} \[0.5em] \text{t}_1 = \dfrac{30 \text{km}}{60 \text{km h}^{-1}} \[0.5em] \Rightarrow \text{t}_1 = \dfrac{1}{2} \text{h} \[0.5em] \Rightarrow \text{t}_1 = 0.5 \text{h} \[0.5em]t1​=v1​S1​​t1​=60km h−130km​⇒t1​=21​h⇒t1​=0.5hConverting 0.5 h to min we get,⇒t1=0.5×60 min⇒t1=30 min\Rightarrow \text{t}_1 = 0.5 \times 60 \text{ min} \[0.5em] \Rightarrow \text{t}_1 = 30 \text{ min} \[0.5em]⇒t1​=0.5×60 min⇒t1​=30 minandt2=S2v2t2=30km40km h−1⇒t2=34h⇒t2=0.75h\text{t}_2 = \dfrac{\text{S}_2}{\text{v}_2} \[0.5em] \text{t}_2 = \dfrac{30 \text{km}}{40 \text{km h}^{-1}} \[0.5em] \Rightarrow \text{t}_2 = \dfrac{3}{4} \text{h} \[0.5em] \Rightarrow \text{t}_2 = 0.75 \text{h} \[0.5em]t2​=v2​S2​​t2​=40km h−130km​⇒t2​=43​h⇒t2​=0.75hConverting 0.75 h to min we get,⇒t2=0.75×60 min⇒t2=45 min\Rightarrow \text{t}_2 = 0.75 \times 60 \text{ min} \[0.5em] \Rightarrow \text{t}_2 = 45 \text{ min} \[0.5em]⇒t2​=0.75×60 min⇒t2​=45 minTotal time t = t1 + t2= 30 min + 45 min= 75 minHence, Total time of journey = 75 min.(ii) As we know,Total distance S = S1 + S2= 30 km + 30 km= 60 kmHence, Total distance travelled = 60 km.As we know,Average speed=total distancetotal time takenAverage speed=60 km1.25 h⇒Average speed=48 kmh−1\text{Average speed} = \dfrac{\text{total distance}}{\text{total time taken}} \[0.5em] \text{Average speed} = \dfrac{60 \text { km}}{1.25 \text { h}} \[0.5em] \Rightarrow \text{Average speed} = 48 \text { kmh} ^{-1}Average speed=total time takentotal distance​Average speed=1.25 h60 km​⇒Average speed=48 kmh−1Hence, Average speed of the car = 48 kmh-1

Question 3

Question 16A car is moving in a straight line with speed 18 km h-1. It is stopped in 5s by applying the brakes.Find —(i) the speed of car in m s -1,(ii) the retardation and(iii) the speed of car after 2 s of applying the brakes .

Accepted Answer

Given,speed of car = 18 km h-1(i) To convert speed to m s-118km h−1=18km1h=18×1000m60×60s⇒18km h−1=18×10m6×6s⇒18km h−1=5m s−118 \text {km h}^{-1} = \dfrac{18 \text {km}}{1 \text{h}} = \dfrac {18 \times 1000 \text{m}}{60 \times 60 \text {s}} \[0.5em] \Rightarrow 18 \text {km h}^{-1} = \dfrac {18 \times 10 \text {m}}{6 \times 6 \text {s}} \[0.5em] \Rightarrow 18 \text {km h}^{-1} = 5 \text {m s}^{-1} \[0.5em]18km h−1=1h18km​=60×60s18×1000m​⇒18km h−1=6×6s18×10m​⇒18km h−1=5m s−1Hence, 18 km h-1 is equal to 5 m s-1.(ii) As the car is stopped, the final velocity (v) = 0initial velocity (u) = 18 km h-1As we know,a=final vel. (v)−initial vel. (u)time (t)\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}a=time (t)final vel. (v)−initial vel. (u)​Given,Initial velocity (u) = 5 m s-1Final velocity (v) = 0Time (t) = 5 sSubstituting the values in the formula above, we get,Acceleration (a)=0−55Acceleration (a)=−55⇒Acceleration (a)=−1ms−2\text {Acceleration (a)} = \dfrac {0 - 5}{5} \[0.5em] \text {Acceleration (a)} = -\dfrac {5}{5} \[0.5em] \Rightarrow \text {Acceleration (a)} = - 1 \text {ms}^{-2} \[0.5em]Acceleration (a)=50−5​Acceleration (a)=−55​⇒Acceleration (a)=−1ms−2Hence, acceleration of the car is -1 m s-2. Negative sign shows that the velocity decreases with time, so retardation is 1 m s -2.(iii) Given,t = 2 sa = -1 m s-2u = 5 m s-1Substituting the values in the formula for acceleration, we get,−1ms−2=(v−5)ms−12s−2ms−1=(v−5)ms−1⇒v=(5−2)ms−1⇒v=3ms−1- 1 \text {ms}^{-2} = \dfrac {(v - 5 ) \text {ms}^{-1}}{2 \text {s}} \[0.5em] - 2 \text {ms}^{-1} = (v - 5) {\text {ms}}^{-1} \[0.5em] \Rightarrow v = (5 - 2) \text {ms}^{-1} \[0.5em] \Rightarrow v = 3 \text {ms}^{-1} \[0.5em]−1ms−2=2s(v−5)ms−1​−2ms−1=(v−5)ms−1⇒v=(5−2)ms−1⇒v=3ms−1Hence, the speed of car after 2 s of applying the brakes = 3 m s-1.Give an example of motion of a body moving with a constant speed, but with a variable velocity. Draw a diagram to represent such a motion.The motion of a body in circular path, is an example of a body moving with a constant speed and variable velocity because the direction of motion of body changes continuously with time.At any instant, the velocity is along the tangent to the circular path at that point.The figure above shows the direction of velocity v at different points A, B, C and D of the circular path.

Question 4

Question 4Figure shows the displacement of a body at different times.(a) Calculate the velocity of the body as it moves for time interval(i) 0 to 5 s,(ii) 5 s to 7 s and(iii) 7 s to 9 s.(b) Calculate the average velocity during the time interval 5 s to 9 s.[ Hint — From 5 s to 9 s, displacement = 7 m - 3 m = 4 m ]

Accepted Answer

Below is the displacement-time graph of the body with the different points marked :(a) As we know,(i) At t = 0 to 5 sVelocity = Slope of straight line OAVelocity=AEOE=(3−0)m(5−0)s⇒Velocity=3m5s⇒Velocity=0.6ms−1\text {Velocity} = \dfrac {\text {AE}}{\text {OE}} = \dfrac {(3 - 0) \text {m}}{(5 - 0) \text {s}} \[0.5em] \Rightarrow \text {Velocity} = \dfrac {3 \text {m}}{5 \text {s}} \[0.5em] \Rightarrow \text {Velocity} = 0.6 {\text {ms}}^{-1} \[0.5em]Velocity=OEAE​=(5−0)s(3−0)m​⇒Velocity=5s3m​⇒Velocity=0.6ms−1Hence, velocity at t = 0 to 5 s = 0.6 m s-1(ii) At t = 5 to 7 sIn this part we observe that there is no change in Y axis, (i.e. displacement is zero so the body is stationary).Hence, velocity at t = 5 to 7 s = 0 m s-1(iii) At t = 7 s to 9 sVelocity = Slope of straight line BCVelocity=CDBD=(7−3)m(9−7)s⇒Velocity=4m2s⇒Velocity=2ms−1\text {Velocity} = \dfrac {\text {CD}}{\text {BD}} = \dfrac {(7 - 3) \text {m}}{(9 - 7) \text {s}} \[0.5em] \Rightarrow \text {Velocity} = \dfrac {4 \text {m}}{2 \text {s}} \[0.5em] \Rightarrow \text {Velocity} = 2 {\text {ms}}^{-1} \[0.5em]Velocity=BDCD​=(9−7)s(7−3)m​⇒Velocity=2s4m​⇒Velocity=2ms−1Hence, velocity at t = 7 s to 9 s = 2 m s-1(b) As we know,Avg. velocity=Total distanceTotal time\text {Avg. velocity} = \dfrac {\text {Total distance}}{\text {Total time}} \[0.5em]Avg. velocity=Total timeTotal distance​Substituting the values from the graph we get,Avg. velocity=(7−3)m(9−5)sAvg. velocity=4m4s⇒Avg. velocity=1ms−1\text {Avg. velocity} = \dfrac {(7 - 3) \text {m}}{(9 - 5)\text {s}} \[0.5em] \text {Avg. velocity} = \dfrac {4 \text {m}}{4\text {s}} \[0.5em] \Rightarrow \text {Avg. velocity} = 1\text {ms} ^ {-1} \[0.5em]Avg. velocity=(9−5)s(7−3)m​Avg. velocity=4s4m​⇒Avg. velocity=1ms−1Hence, average velocity of the car is 1 ms-1.

Question 5

Question 9Figure below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD.(i) State how is the distance travelled in any part determined from this graph.(ii) Compare the distance travelled in part BC with the distance travelled in part AB.(iii) Which part of graph shows motion with uniform (a) velocity (b) acceleration (c) retardation?(iv) (a) Is the magnitude of acceleration higher or lower than that of retardation? Give a reason. (b) Compare the magnitude of acceleration and retardation.

Accepted Answer

(i) The distance travelled in any part of the graph can be obtained by finding the area enclosed by the graph in that part with the time axis.(ii) Let E be the point on time axis (x axis) at time t and F be the point at time 2t.When we observe the graph, we find,Distance travelled in part BC = Area of rectangle EBCF∴ Distance travelled in part BC = length x breadth= (2t - t) x (v0 - 0)= t x v0     .... [1]Distance travelled in part AB = Area of triangle ABT∴ Distance travelled in part AB = 12\dfrac{1}{2}21​ x base x height= 12\dfrac{1}{2}21​ x (t - 0) x (v0 - 0)= 12\dfrac{1}{2}21​ x t x v0     .... [2]Comparing [1] and [2] we get,Distance travelled in part BC : Distance travelled in part AB=t v0tv02=21= \dfrac {\text {t v}_0}{\dfrac{\text {tv}_0}{2}} \[0.5em] = \dfrac {2}{1}=2tv0​​t v0​​=12​Hence,Distance travelled in part BC : Distance travelled in part AB = 2 : 1(iii) The different parts of the graph are mentioned below:(a) Uniform velocity is shown in part BC of the graph, as the velocity is constant with time.(b) Uniform acceleration is shown in part AB of the graph, as the velocity is increasing with time.(c) Uniform retardation is shown in part CD of the graph, as the velocity is decreasing with time.(iv) (a) The magnitude of acceleration is lower, as slope of line AB is less than that of line CD.(b) Acceleration in part AB = slope of ABSlope of line AB=v0−0t−0⇒Slope of line AB=v0t\text {Slope of line AB} = \dfrac{\text v_0 - 0}{\text t - 0} \[0.5em] \Rightarrow\text {Slope of line AB} = \dfrac{\text v_0}{\text t} \[0.5em]Slope of line AB=t−0v0​−0​⇒Slope of line AB=tv0​​Retardation in part CD = slope of CDSlope of line CD=v0−02.5t−2t⇒Slope of line AB=v00.5t\text {Slope of line CD} = \dfrac{\text v_0 - 0}{2.5\text t - 2\text t} \[0.5em] \Rightarrow\text {Slope of line AB} = \dfrac{\text v_0}{0.5\text t} \[0.5em]Slope of line CD=2.5t−2tv0​−0​⇒Slope of line AB=0.5tv0​​Magnitude of acceleration : Magnitude of retardation = Slope of line AB : Slope of line CD=v0tv00.5t=0.51=510=12= \dfrac{\dfrac{\text v_0}{\text t}}{\dfrac{\text v_0}{0.5\text t}} \[1em] = \dfrac{0.5}{1} \[1em] = \dfrac{5}{10} \[1em] = \dfrac{1}{2}=0.5tv0​​tv0​​​=10.5​=105​=21​Hence, Magnitude of acceleration : Magnitude of retardation = 1 : 2

Question 6

Question 10The velocity-time graph of a moving body is given below in figureFind —(i) The acceleration in parts AB, BC and CD.(ii) Displacement in each part AB, BC, CD, and(iii) Total displacement.

Accepted Answer

Let E be the point at t = 4 and F be the point at t = 8 as labelled in the graph below :(i) Acceleration in part AB = slope of ABslope of AB=(30−0)ms−1(4−0)sslope of AB=30ms−14sslope of AB=7.5ms−2\text {slope of AB} = \dfrac{(30 - 0) \text {ms}^{-1}}{(4-0) \text {s}} \[0.5em] \text {slope of AB} = \dfrac{30 \text {ms}^{-1}}{4 \text {s}} \[0.5em] \text {slope of AB} = 7.5 \text {ms}^{-2} \[0.5em]slope of AB=(4−0)s(30−0)ms−1​slope of AB=4s30ms−1​slope of AB=7.5ms−2Hence, Acceleration in part AB = 7.5 m s-2Acceleration in part BC = slope of BCWe observe from the graph that there is no change in velocity in part BC. Hence, Acceleration in part BC = 0 m s-1Acceleration in part CD = slope of CDslope of CD=(0−30)ms−1(10−8)sslope of CD=−30ms−12sslope of CD=−15ms−2\text {slope of CD} = \dfrac{(0 - 30) \text {ms}^{-1}}{(10-8) \text {s}} \[0.5em] \text {slope of CD} = \dfrac{-30 \text {ms}^{-1}}{2 \text {s}} \[0.5em] \text {slope of CD} = -15 \text {ms}^{-2} \[0.5em]slope of CD=(10−8)s(0−30)ms−1​slope of CD=2s−30ms−1​slope of CD=−15ms−2Hence, Acceleration in part CD = -15 m s-1(ii) Displacement in each part is as follows —(a) Displacement of part AB = Area of triangle ABE=12×base×height= \dfrac {1}{2} \times \text {base} \times \text {height}=21​×base×heightSubstituting the values in the formula above, we get,=12×(4−0)s×(30−0)m s−1=12×4s×30m s−1=30m= \dfrac {1}{2} \times {(4 -0) \text {s}} \times (30 - 0)\text {m s}^{-1} \[0.5em] = \dfrac {1}{2} \times 4 \text {s} \times 30 \text {m s}^{-1} \[0.5em] = 30 \text {m} \[0.5em]=21​×(4−0)s×(30−0)m s−1=21​×4s×30m s−1=30mHence,Displacement of part AB = 60 m(b) Displacement of part BC = Area of Square EBCF=length×breadth= \text {length} \times \text {breadth}=length×breadthSubstituting the values in the formula above, we get,=(8−4)s×(30−0)m s−1=4×30m=120m= {(8 -4) \text {s}} \times (30 - 0)\text {m s}^{-1} \[0.5em] = 4 \times 30 \text {m} \[0.5em] = 120 \text {m} \[0.5em]=(8−4)s×(30−0)m s−1=4×30m=120mHence,Displacement of part BC = 120 m(c) Displacement of part CD = Area of triangle CDF=12×base×height= \dfrac {1}{2} \times \text {base} \times \text {height}=21​×base×heightSubstituting the values in the formula above, we get,=12×(10−8)s×(30−0)m s−1=12×2×30m=30m= \dfrac {1}{2} \times {(10 -8) \text {s}} \times (30 - 0)\text {m s}^{-1} \[0.5em] = \dfrac {1}{2} \times 2 \times 30 \text {m} \[0.5em] = 30 \text {m} \[0.5em]=21​×(10−8)s×(30−0)m s−1=21​×2×30m=30mHence,Displacement of part CD = 30 m(iii) Total displacement = Displacement of part AB + Displacement of part BC + Displacement of part CD= 60 + 30 + 120 = 210Hence,total displacement = 210 m

Question 7

Question 12Figure shows the velocity-time graph of a particle moving in a straight line.(i) State the nature of motion of particle.(ii) Find the displacement of particle at t = 6 s.(iii) Does the particle change its direction of motion?(iv) Compare the distance travelled by the particle from 0 to 4 s and from 4 s to 6 s.(v) Find the acceleration from 0 to 4 s and retardation from 4 s to 6 s.

Accepted Answer

(i) As we observe the graph, we find that, the nature of motion of particle is that, the particles are uniformly accelerated from 0 to 4s and then uniformly retarded from 4s to 6s.(ii) As we know,displacement of particles can be obtained by finding the area enclosed by the graph in that part with the time axis up to that instance.At t = 6 s,Displacement = area of triangle=12×base×height= \dfrac {1}{2} \times \text {base} \times \text {height}=21​×base×heightSubstituting the values in the formula above, we get,=12×(6−0)s×(2−0)m s−1=12×6×2m=6m= \dfrac {1}{2} \times {(6 - 0) \text {s}} \times (2 - 0)\text {m s}^{-1} \[0.5em] = \dfrac {1}{2} \times 6 \times 2 \text {m} \[0.5em] = 6 \text {m} \[0.5em]=21​×(6−0)s×(2−0)m s−1=21​×6×2m=6mHence,Displacement of particle at t = 6 s is 6 m(iii) No, the particle does not change its direction of motion.(iv) At t = 0 to 4 s,Distance covered = area of triangle=12×base×height= \dfrac {1}{2} \times \text {base} \times \text {height}=21​×base×heightSubstituting the values in the formula above, we get,=12×(4−0)s×(2−0)m s−1=12×4s×2m s−1=4m= \dfrac {1}{2} \times {(4 - 0) \text {s}} \times (2 - 0)\text {m s}^{-1} \[0.5em] = \dfrac {1}{2} \times 4 \text {s} \times 2 \text {m s}^{-1} \[0.5em] = 4 \text {m} \[0.5em]=21​×(4−0)s×(2−0)m s−1=21​×4s×2m s−1=4mHence,Distance covered between 0 to 4 s = 4 mAt t = 4 s to 6 s,Distance covered = area of triangle=12×base×height= \dfrac {1}{2} \times \text {base} \times \text {height}=21​×base×heightSubstituting the values in the formula above, we get,=12×(6−4)s×(2−0)m s−1=12×2s×2m s−1=2m= \dfrac {1}{2} \times {(6 - 4) \text {s}} \times (2 - 0)\text {m s}^{-1} \[0.5em] = \dfrac {1}{2} \times 2 \text {s} \times 2 \text {m s}^{-1} \[0.5em] = 2 \text {m} \[0.5em]=21​×(6−4)s×(2−0)m s−1=21​×2s×2m s−1=2mHence,Distance covered between 4s to 6 s = 2 m∴ Distance covered between 0 to 4 s : Distance covered between 4 s to 6 s = 4 : 2 = 2 : 1(v) Acceleration in part 0 s to 4 s = slope of graphslope=(2−0)ms−1(4−0)sslope=2ms−14sslope=0.5ms−2\text {slope} = \dfrac{(2 - 0) \text {ms}^{-1}}{(4-0) \text {s}} \[0.5em] \text {slope} = \dfrac{2 \text {ms}^{-1}}{4 \text {s}} \[0.5em] \text {slope} = 0.5 \text {ms}^{-2} \[0.5em]slope=(4−0)s(2−0)ms−1​slope=4s2ms−1​slope=0.5ms−2Hence,Acceleration in part 0 s to 4 s = 0.5 m s-2As we know,Retardation in part 4 s to 6 s = slope of graphslope=(0−2)ms−1(6−4)sslope=−2ms−12sslope=−1ms−2\text {slope} = \dfrac{(0 - 2) \text {ms}^{-1}}{(6 - 4) \text {s}} \[0.5em] \text {slope} = \dfrac{- 2 \text {ms}^{-1}}{2\text {s}} \[0.5em] \text {slope} = - 1 \text {ms}^{-2} \[0.5em]slope=(6−4)s(0−2)ms−1​slope=2s−2ms−1​slope=−1ms−2Acceleration = -1 ms-2 and as we know retardation is negative accelerationHence, Retardation in part 4 s to 6 s = 1 m s-2

Question 8

Question 17A body falls freely from a certain height. Show graphically the relation between the distance fallen and square of time. How will you determine g from this graph?

Accepted Answer

Below is the graph of distance fallen against square of time for a freely falling body :The slope is half the acceleration due to gravity.Thus, the value of acceleration due to gravity (g) can be obtained by doubling the slope of the S - t2 graph for a freely falling body.

Question 9

Question 1Derive the following equations for a uniformly accelerated motion —(i) v = u + at(ii) S = ut + 12\dfrac{1}{2}21​ at2(iii) v2 = u2 + 2aSwhere the symbols have their usual meanings.

Accepted Answer

(i) By definition,Acceleration=change in velocitytime taken⇒Acceleration=final velocity - initial velocitytime taken\text {Acceleration} = \dfrac {\text {change in velocity}}{\text {time taken}} \[0.5em] \Rightarrow \text {Acceleration} = \dfrac {\text {final velocity - initial velocity}}{\text {time taken}} \[0.5em]Acceleration=time takenchange in velocity​⇒Acceleration=time takenfinal velocity - initial velocity​Hence,a=v−utat=v−u⇒v=u+at\text {a} = \dfrac{\text{v} - \text{u}}{\text {t}} \[0.5em] \text {at} = \text{v} - \text{u} \[0.5em] \Rightarrow \text {v} = \text{u} + \text{at} \[0.5em]a=tv−u​at=v−u⇒v=u+atHence, we get,v = u + at     [Eq. 1](ii) By definition,Distance travelled=Avg. Velocity×timeDistance travelled=(Initial Vel.+Final Vel.2)×time\text {Distance travelled} = \text {Avg. Velocity} \times \text{time} \[1em] \text {Distance travelled} = \Big(\dfrac{\text {Initial Vel.} + \text {Final Vel.}}{2}\Big) \times \text{time}Distance travelled=Avg. Velocity×timeDistance travelled=(2Initial Vel.+Final Vel.​)×timeorS=(u+v2)×t\text{S} = \Big(\dfrac {\text{u} + \text{v}}{2}\Big) \times \text{t}S=(2u+v​)×tBut, from Eq. 1, v = u + atTherefore,S=(u + ( u +at)2)×tS=(2u + at2)×t⇒S=ut+12at2\text{S} = \Big(\dfrac{\text {u + ( u +at)}}{2}\Big) \times \text {t} \[0.5em] \text{S} = \Big(\dfrac{\text {2u + at}}{2}\Big) \times \text {t} \[0.5em] \Rightarrow \text{S} = \text {ut} + \dfrac{1}{2} \text {at}^2 \[0.5em]S=(2u + ( u +at)​)×tS=(22u + at​)×t⇒S=ut+21​at2Hence, we get,S = ut + 12\bold{\dfrac{1}{2}}21​ at2(iii) By definition,Distance travelled=Avg. Velocity×time⇒S=(u+v2)×t\text {Distance travelled} = \text {Avg. Velocity} \times \text{time} \[0.5em] \Rightarrow \text{S} = \Big(\dfrac {\text{u} + \text{v}}{2}\Big) \times \text{t}Distance travelled=Avg. Velocity×time⇒S=(2u+v​)×tBut, from Eq. 1, v = u + ator t = v−ua\dfrac{\text{v}-\text{u}}{\text{a}}av−u​Therefore,S=(v+u2)×(v−ua)⇒S=v2−u22a\text{S} = \Big(\dfrac {\text{v} + \text{u}}{2}\Big) \times \Big(\dfrac{\text{v}-\text{u}}{\text a}\Big) \[1em] \Rightarrow \text{S} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}S=(2v+u​)×(av−u​)⇒S=2av2−u2​So, v2 - u2 = 2aSHence,v2 = u2 + 2aS

Question 10

Question 1(vii)A car starting from rest moves on a straight path for time t = 0 to t = T with a uniform acceleration a and then stops with a uniform retardation. The average speed of the car will be :aT4\dfrac{	ext {aT}}{4}4aT​3aT2\dfrac{3	ext {aT}}{2}23aT​aT2\dfrac{	ext {aT}}{2}2aT​aT

Accepted Answer

aT2\dfrac{\text {aT}}{2}2aT​ReasonAssuming, uniform retardation has the same value as the uniform acceleration (a).Case 1 : When car is acceleratingGiven,Initial speed (u1) = 0Acceleration (a1) = aTime (t1) = TLet, distance travelled be S1 and final velocity is v1.As,v = u + atOn putting valuesv1 = u1 + a1t1 = 0 + aT = aTBy using,S = ut + 12\dfrac{1}{2}21​at2On putting valuesS1 = u1t1 + 12\dfrac{1}{2}21​a1t12\text t_1^2t12​ = 0 + 12\dfrac{1}{2}21​aT2 = 12\dfrac{1}{2}21​aT2Case 2 : When car is retardingGiven,Initial speed (u2) = v1 = aTAcceleration (a2) = -aFinal speed (v2) = 0Let, distance travelled be S2 and time taken is t2.As,v = u + atOn putting valuesv2 = u2 + a2t2⟹ 0 = aT - at2⟹ aT = at2⟹ t2 = TBy using,S = ut + 12\dfrac{1}{2}21​at2On putting valuesS2 = u2t2 + 12\dfrac{1}{2}21​a2t22\text t_2^2t22​= (aT)(T) - 12\dfrac{1}{2}21​aT2= aT2 - 12\dfrac{1}{2}21​aT2= 12\dfrac{1}{2}21​aT2Here,Total distance travelled (S) = S1 + S2 = 12\dfrac{1}{2}21​aT2 + 12\dfrac{1}{2}21​aT2 = aT2Total time taken (t) = t1 + t2 = T + T = 2TThen,Avg. Speed (v)=Total distance (S)Total time (t)=aT22T=aT2\text {Avg. Speed (v)}=\dfrac{\text {Total distance (S)}}{\text {Total time (t)}} \[1em] =\dfrac{\text {aT}^2}{2\text T}= \dfrac{\text {aT}}{2}Avg. Speed (v)=Total time (t)Total distance (S)​=2TaT2​=2aT​∴ Average speed of the car is aT2.\dfrac{ {\bold {aT}}}{\bold 2}.2aT​.

Question 11

Question 2A body starts with an initial velocity of 10 m s-1 and acceleration 5 m s-2. Find the distance covered by it in 5 s .

Accepted Answer

According to equation of motion —S = ut + 12\dfrac{1}{2}21​ at2Where,distance covered = Sinitial velocity u = 10 m s-1acceleration a = 5 m s-2time t = 5 sSubstituting the values in the formula, we get,S=(10×5)+(12×5×52)S=(50)+(12×5×25)S=(50)+(2.5×25)S=50+62.5⇒S=112.5 m\text S = (10 \times 5) + (\dfrac{1}{2} \times 5 \times 5^2) \[0.5em] \text S = (50) + (\dfrac{1}{2} \times 5 \times 25) \[0.5em] \text S = (50) + ( 2.5 \times 25) \[0.5em] \text S = 50 + 62.5 \[0.5em] \Rightarrow \text S = 112.5\ \text m \[0.5em]S=(10×5)+(21​×5×52)S=(50)+(21​×5×25)S=(50)+(2.5×25)S=50+62.5⇒S=112.5 mHence, distance covered = 112.5 m.

Question 12

Question 3A vehicle is accelerating on a straight road. Its velocity at any instant is 30 km h-1, after 2 s, it is 33.6 km h-1 and after further 2 s, it is 37.2 km h-1. Find the acceleration of vehicle in m s-2. Is the acceleration uniform ?

Accepted Answer

As we know,a = v−ut\dfrac{\text v - \text u}{\text t}tv−u​After t = 2s,u = 30 km h-1Converting km h-1 to m s-1We get,30km h−1=30km1h=30×1000m60×60s⇒30km h−1=30×10m6×6s⇒30km h−1=8.33×ms30 \text{km h}^{-1} = \dfrac{30 \text{km}}{1\text{h}} = \dfrac {30 \times 1000 \text{m}}{60 \times 60 \text{s}} \[0.5em] \Rightarrow 30 \text{km h}^{-1} = \dfrac {30 \times 10 \text{m}}{6 \times 6 \text{s}} \[0.5em] \Rightarrow 30 \text{km h}^{-1} = \dfrac {8.33 \times \text{m}}{\text{s}} \[0.5em]30km h−1=1h30km​=60×60s30×1000m​⇒30km h−1=6×6s30×10m​⇒30km h−1=s8.33×m​Hence, 30 km h-1 is equal to 8.33 m s-1.Now,v = 33.6 km h-1Converting km h-1 to m s-1We get,33.6km h−1=33.6km1h=33.6×1000m60×60s⇒33.6km h−1=33.6×10m6×6s⇒33.6km h−1=9.33×ms33.6 \text{km h}^{-1} = \dfrac{33.6 \text{km}}{1\text{h}} = \dfrac {33.6 \times 1000 \text{m}}{60 \times 60 \text{s}} \[0.5em] \Rightarrow 33.6 \text{km h}^{-1} = \dfrac {33.6 \times 10 \text{m}}{6 \times 6 \text{s}} \[0.5em] \Rightarrow 33.6 \text{km h}^{-1} = \dfrac {9.33 \times \text{m}}{\text{s}} \[0.5em]33.6km h−1=1h33.6km​=60×60s33.6×1000m​⇒33.6km h−1=6×6s33.6×10m​⇒33.6km h−1=s9.33×m​Hence, 33.6 km h-1 is equal to 9.33 ms-1.Substituting the values in the formula above, we get,a=9.33−8.332a=12⇒a=0.5 m s2\text a = \dfrac{9.33 - 8.33}{2} \[0.5em] \text a = \dfrac{1}{2} \[0.5em] \Rightarrow a = 0.5\ \text {m s}^{2} \[0.5em]a=29.33−8.33​a=21​⇒a=0.5 m s2Hence, acceleration in the first 2 s = 0.5 m s -2For the next 2 s,u = 9.33 m s-1v = 37.2 km h-1Converting km h-1 to m s-1We get,37.2km h−1=37.2km1h=37.2×1000m60×60s⇒37.2km h−1=37.2×10m6×6s⇒37.2km h−1=10.33×ms37.2 \text{km h}^{-1} = \dfrac{37.2\text{km}}{1\text{h}} = \dfrac {37.2 \times 1000 \text{m}}{60 \times 60 \text{s}} \[0.5em] \Rightarrow 37.2 \text{km h}^{-1} = \dfrac {37.2 \times 10 \text{m}}{6 \times 6 \text{s}} \[0.5em] \Rightarrow 37.2 \text{km h}^{-1} = \dfrac {10.33 \times \text{m}}{\text{s}} \[0.5em]37.2km h−1=1h37.2km​=60×60s37.2×1000m​⇒37.2km h−1=6×6s37.2×10m​⇒37.2km h−1=s10.33×m​Hence, 37.2 km h-1 is equal to 10.33 ms-1.Substituting the values in the formula above, we get,a=10.33−9.332a=12⇒a=0.5 m s2\text a = \dfrac{10.33 - 9.33 }{2} \[0.5em] \text a = \dfrac{1}{2} \[0.5em] \Rightarrow \text a = 0.5\ \text {m s}^{2} \[0.5em]a=210.33−9.33​a=21​⇒a=0.5 m s2Hence, acceleration is 0.5 m s -2Yes, the acceleration is uniform as the acceleration in both instances is same.

Question 13

Question 5A bullet initially moving with a velocity 20 m s-1 strikes a target and comes to rest after penetrating a distance 10 cm in the target. Calculate the retardation caused by the target.

Accepted Answer

According to equation of motion —v2 = u2 + 2aSGiven,u = 20 m s-1,v = 0,s = 10 cm = 0.1 mSubstituting the values in the formula above, we get,0=(20)2+(2×a×0.1)⇒0=400+0.2a⇒−400=0.2a⇒a=−4000.2⇒a=−40002⇒a=−2000 ms−20 = (20)^2 + (2\times \text a \times 0.1) \[0.5em] \Rightarrow 0 = 400 + 0.2 \text a \[0.5em] \Rightarrow -400 = 0.2\text a \[0.5em] \Rightarrow \text a = -\dfrac{400}{0.2} \[0.5em] \Rightarrow \text a = -\dfrac{4000}{2} \[0.5em] \Rightarrow \text a = - 2000\ \text {ms}^{-2}\[0.5em]0=(20)2+(2×a×0.1)⇒0=400+0.2a⇒−400=0.2a⇒a=−0.2400​⇒a=−24000​⇒a=−2000 ms−2Hence, Retardation is 2000 m s-2

Question 14

Question 8A train is moving with a velocity of 90 km h-1. It is brought to stop by applying the brakes which produce a retardation of 0.5 m s-2.Find —(i) the velocity after 10 s, and(ii) the time taken by the train to come to rest.

Accepted Answer

(i) As we know, according to the equation of motion,v = u + atGiven,u = 90 km h-1velocity in m s-190km h−1=90km1h=90×1000m60×60s⇒90km h−1=90×10m6×6s⇒90km h−1=15×10m1×6s⇒90km h−1=25×ms90 \text{km h}^{-1} = \dfrac{90 \text{km}}{1\text{h}} = \dfrac {90 \times 1000 \text{m}}{60 \times 60 \text{s}} \[0.5em] \Rightarrow 90 \text{km h}^{-1} = \dfrac {90 \times 10 \text{m}}{6 \times 6 \text{s}} \[0.5em] \Rightarrow90 \text{km h}^{-1} = \dfrac {15 \times 10 \text{m}}{1 \times 6 \text{s}} \[0.5em] \Rightarrow 90 \text{km h}^{-1} = \dfrac {25 \times \text{m}}{\text{s}} \[0.5em]90km h−1=1h90km​=60×60s90×1000m​⇒90km h−1=6×6s90×10m​⇒90km h−1=1×6s15×10m​⇒90km h−1=s25×m​Hence, 90 km h-1 is equal to 25 m s-1.v = 0 (as the train stops on application of brakes)t = 10 sretardation = -a = -0.5 m s-2Substituting the values in the formula above we get,v=25+[(−0.5)×10]v=25+[−5]⇒v=20m s−1\text {v} = 25 + [(-0.5) \times 10] \[0.5em] \text {v} = 25 + [- 5] \[0.5em] \Rightarrow \text {v} = 20 \text {m s}^{-1} \[0.5em]v=25+[(−0.5)×10]v=25+[−5]⇒v=20m s−1Hence, the velocity after 10 s = 20 m s-1.(ii) As we know, according to the equation of motion,v = u + atSubstituting the values in the formula above, to get the time taken by the train to come to rest.0=25+(−0.5×t)0=25−0.5t0.5t=25⇒t=250.5⇒t=2505⇒t=50 s\text {0} = 25 + (-0.5 \times \text{t}) \[0.5em] \text {0} = 25 - 0.5 \text{t} \[0.5em] 0.5 \text{t} = 25 \[0.5em] \Rightarrow \text {t} = \dfrac {25}{0.5} \[0.5em] \Rightarrow \text {t} = \dfrac {250}{5} \[0.5em] \Rightarrow \text {t} = 50\ \text {s} \[0.5em]0=25+(−0.5×t)0=25−0.5t0.5t=25⇒t=0.525​⇒t=5250​⇒t=50 sHence, the time taken by the train to come to rest = 50 s.

Question 15

Question 11A body moves from rest with a uniform acceleration and travels 270 m in 3 s. Find the velocity of the body at 10 s after the start.

Accepted Answer

As we know,S = ut + 12\dfrac{1}{2}21​ at2Given,u = 0 m s -1S = 270 mt = 3 sSubstituting the values in the formula above we get,270=(0×3)+(12×a×32)270=92×a270×2=9×a⇒a=5409⇒a=60 ms−2270 = (0 \times 3) + (\dfrac{1}{2} \times \text a \times 3^2) \[0.5em] 270 = \dfrac{9}{2} \times \text a \[0.5em] 270 \times 2 = 9 \times \text a \[0.5em] \Rightarrow \text a = \dfrac{540}{9} \[0.5em] \Rightarrow \text a = 60\ \text{ms}^{-2}270=(0×3)+(21​×a×32)270=29​×a270×2=9×a⇒a=9540​⇒a=60 ms−2Hence, a = 60 m s-2To find out velocity after t = 10 s.v = u + atSubstituting the values in the formula above we get,v=0+(60×10)⇒v=600 ms−1\text v = 0 + (60 \times 10) \[0.5em] \Rightarrow \text v = 600\ \text{ms}^{-1} \[0.5em]v=0+(60×10)⇒v=600 ms−1Hence, velocity after 10 s = 600 m s-1

Question 16

Question 13A car travels with a uniform velocity of 25 m s-1 for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.Find —(i) the distance which the car travels before the brakes are applied,(ii) the retardation and(iii) the distance travelled by the car after applying the brakes.

Accepted Answer

(i) As we know,Distance = Speed x timeInitial velocity = u = 25 m s-1Final velocity = 0time = 5 sSubstituting the values in the formula above we get,Distance=25×5⇒Distance=125 m\text{Distance} = 25 \times 5 \[0.5em] \Rightarrow \text{Distance} = 125\ mDistance=25×5⇒Distance=125 mHence, distance covered = 125 m.(ii) Retardation = -a = v−ut\dfrac{\text v - \text u}{\text t}tv−u​t = 10 sSubstituting the values in the formula above we get,Retardation=−0−2510⇒Retardation=2510⇒Retardation=2.5 m s−2\text {Retardation} = -\dfrac{0 - 25}{10} \[0.5em] \Rightarrow \text {Retardation} = \dfrac{25}{10} \[0.5em] \Rightarrow \text {Retardation} = 2.5\ \text {m s}^{-2} \[0.5em]Retardation=−100−25​⇒Retardation=1025​⇒Retardation=2.5 m s−2Hence, retardation of the car = 2.5 m s-2(iii) As we know,v2 - u2 = 2aSSubstituting the values in the formula above we get,(0)2−(25)2=2×(−2.5)×S0−625=−5S⇒−5S=−625 ⇒S=6255⇒S=125 m(0)^2 - (25)^2 = 2 \times (-2.5) \times \text S \[0.5em] 0 - 625 = - 5\text S \[0.5em] \Rightarrow - 5\text S = - 625\ \[0.5em] \Rightarrow \text S = \dfrac{625}{5} \ \[0.5em] \Rightarrow \text S = 125\ \text m \ \[0.5em](0)2−(25)2=2×(−2.5)×S0−625=−5S⇒−5S=−625 ⇒S=5625​⇒S=125 mHence, the distance travelled by the car after applying the brakes = 125 m.

Question 17

Question 14A space craft flying in a straight course with a velocity of 75 km s-1 fires its rocket motors for 6.0 s. At the end of this time, its speed is 120 km s-1 in the same direction.Find —(i) the space craft’s average acceleration while the motors were firing(ii) the distance travelled by the space craft in the first 10 s after the rocket motors were started, the motors having been in action for only 6.0 s.

Accepted Answer

(i) As we know,v - u = atGiven,u = 75 km s-1v = 120 km s-1t = 6 sSubstituting the values in the formula above we get,120−75=a×645=6a⇒a=456⇒a=7.5 km s−2120 - 75 = \text a \times 6 \[0.5em] 45 = 6\text a \[0.5em] \Rightarrow \text a = \dfrac{45}{6} \[0.5em] \Rightarrow \text a = 7.5 \text { km s}^{-2} \[0.5em]120−75=a×645=6a⇒a=645​⇒a=7.5 km s−2Hence, acceleration = 7.5 km s-2(ii) As we know,S = ut + 12\dfrac{1}{2}21​at2For the first 6 s,u = 75 km s-1a = 7.5 km s-2S1=(75×6)+(12×7.5×62)S1=450+(12×7.5×36)S1=450+(7.5×18)S1=450+135⇒S1=585 km\text S_{1} = (75 \times 6) + (\dfrac{1}{2} \times 7.5 \times 6^2) \[0.5em] \text S_{1} = 450 + (\dfrac{1}{2} \times 7.5 \times 36) \[0.5em] \text S_{1} = 450 + (7.5 \times 18) \[0.5em] \text S_{1} = 450 + 135 \[0.5em] \Rightarrow \text S_{1} = 585 \text { km}S1​=(75×6)+(21​×7.5×62)S1​=450+(21​×7.5×36)S1​=450+(7.5×18)S1​=450+135⇒S1​=585 kmHence, S1 = 585 kmFor the next 4 sS2 = speed x timeGiven,t = 4 sspeed = 120 km s -1Substituting the values in the formula above, we get,S2=120×4S2=480 km\text S_2 = 120 \times 4 \[0.5em] \text S_2 = 480 \text { km}S2​=120×4S2​=480 kmHence, S2 = 480 kmTotal distance covered by the aircraft = S1 + S2= 585 + 480= 1065 kmHence,Total distance covered by the aircraft = 1065 km.

Motion in One Dimension

Exercise 2A Multiple Choice Type

Exercise 2A Numericals

Exercise 2B Numericals

Exercise 2B Short Answer Type

Exercise 2C Long Answer Type

Exercise 2C Multiple Choice Type

Exercise 2C Numericals