The figure shows a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion,give the signs of $v$ and $a$ in the three intervals. What are the accelerations at the points $A, B, C$ and $D$?

(N/A) $1$. Average acceleration is greatest in interval $2$.
$2$. Average speed is greatest in interval $3$.
$3$. $v$ is positive in intervals $1, 2,$ and $3$. $a$ is positive in interval $1$,negative in interval $2$,and zero in interval $3$.
$4$. Acceleration is zero at points $A, B, C,$ and $D$.
Detailed Explanation:
- Acceleration is given by the slope of the speed-time graph. Since the magnitude of the slope of the speed-time graph is maximum in interval $2$,the average acceleration is greatest in this interval.
- The average speed is represented by the average height of the curve from the time-axis. It is clear that the average height is greatest in interval $3$. Hence,the average speed of the particle is greatest in interval $3$.
- In interval $1$: The slope of the speed-time graph is positive,so acceleration $a$ is positive. The speed $v$ is positive.
- In interval $2$: The slope of the speed-time graph is negative,so acceleration $a$ is negative. The speed $v$ is positive.
- In interval $3$: The speed-time graph is horizontal (slope is zero),so acceleration $a$ is zero. The speed $v$ is positive.
- At points $A, B, C,$ and $D$,the tangent to the curve is parallel to the time-axis,meaning the slope is zero. Therefore,the acceleration at these points is zero.