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Figure gives 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 $D$?
Solution
Average acceleration is greatest in interval 2 Average speed is greatest in interval 3 v is positive in intervals $1,2,$ and $3 a$ is positive in intervals 1 and 3 and negative in interval $2 a=0$ at $A, B, C, D$
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time. since the slope of the given speed-time graph is maximum in interval $2,$ average acceleration will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval $3 .$ Hence, average speed of the particle is the greatest in interval $3$
In interval 1:
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
In interval 2:
The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.
In interval 3:
The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval. Points $A, B, C,$ and $D$ are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points $A , B , C ,$ and $D ,$ acceleration of the particle is zero.
