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ICSE Class 9 Physics Question 17 of 17

Motion in One Dimension — Question 14

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Question

Question 14

A 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.

Answer

(i) As we know,

v - u = at

Given,

u = 75 km s-1

v = 120 km s-1

t = 6 s

Substituting 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]

Hence, acceleration = 7.5 km s-2

(ii) As we know,

S = ut + 12\dfrac{1}{2}at2

For the first 6 s,

u = 75 km s-1

a = 7.5 km s-2

S1=(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}

Hence, S1 = 585 km

For the next 4 s

S2 = speed x time

Given,

t = 4 s

speed = 120 km s -1

Substituting 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}

Hence, S2 = 480 km

Total distance covered by the aircraft = S1 + S2
= 585 + 480
= 1065 km

Hence,
Total distance covered by the aircraft = 1065 km.