Study the speed-time graph of a car below and answer the following questions:
$(a)$ What type of motion is represented by $OA$?
$(b)$ Find the acceleration from $B$ to $C$.
$(c)$ Calculate the distance covered by the body from $A$ to $B$.

(N/A) The segment $OA$ represents uniform acceleration,as the speed increases linearly with time.
$(b)$ Acceleration is given by the slope of the speed-time graph. For the segment $BC$,the initial speed at $B$ is $60 \ m/s$ at $t = 60 \ s$,and the final speed at $C$ is $0 \ m/s$ at $t = 80 \ s$.
$a = \frac{v - u}{t_2 - t_1} = \frac{0 - 60}{80 - 60} = \frac{-60}{20} = -3 \ m/s^2$.
Thus,the acceleration is $-3 \ m/s^2$ (retardation).
$(c)$ The distance covered is equal to the area under the speed-time graph. For the segment $AB$,the motion is at a constant speed of $60 \ m/s$ from $t = 20 \ s$ to $t = 60 \ s$.
Distance = $\text{Speed} \times \text{Time} = 60 \ m/s \times (60 - 20) \ s = 60 \times 40 = 2400 \ m$.