Figure 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 ]

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.