Motion in One Dimension — Question 4

Below is the displacement-time graph of the body with the different points marked :

(a) As we know,

(i) At t = 0 to 5 s

Velocity = Slope of straight line OA

$\text {Velocity} = \dfrac {\text {AE}}{\text {OE}} = \dfrac {(3 - 0) \text {m}}{(5 - 0) \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = \dfrac {3 \text {m}}{5 \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = 0.6 {\text {ms}}^{-1} \\[0.5em]$

Hence, velocity at t = 0 to 5 s = 0.6 m s^-1

(ii) At t = 5 to 7 s

In 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 s

Velocity = Slope of straight line BC

$\text {Velocity} = \dfrac {\text {CD}}{\text {BD}} = \dfrac {(7 - 3) \text {m}}{(9 - 7) \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = \dfrac {4 \text {m}}{2 \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = 2 {\text {ms}}^{-1} \\[0.5em]$

Hence, velocity at t = 7 s to 9 s = 2 m s^-1

(b) As we know,

$\text {Avg. velocity} = \dfrac {\text {Total distance}}{\text {Total time}} \\[0.5em]$

Substituting the values from the graph we get,

$\text {Avg. velocity} = \dfrac {(7 - 3) \text {m}}{(9 - 5)\text {s}} \\[0.5em] \text {Avg. velocity} = \dfrac {4 \text {m}}{4\text {s}} \\[0.5em] \Rightarrow \text {Avg. velocity} = 1\text {ms} ^ {-1} \\[0.5em]$

Hence, average velocity of the car is 1 ms^-1.