Fig. 2 (NCT). 10 . Give a speed-tiem graph of a particle in one dimenstional motion. Three different equal intervals of time are shwon. In which interval is the average acceleration graeatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of `u` and `a` in the three intervals. What are the acclerations at teh points ` A, B , C` and ` D` ?
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Explanation

We know that average acceleration in a small interval of time is equal to slope of velocity-time graph in that interval .as the slope of velocity-time graph is maximum in interval `2` as compared to other intervals ` 1` and `3`, hence the magnitude of average accelration is greatest in interval `2`.
The average speed is greatest in interval `3` for obvious reasons.
In interval `1`, the slope of verlocity-time graph is positive, hence accelration a is positive, the speed (u) is positive in this interval due to obvious reasons.
In niterval `2`, the slope of velocity-tiem graph is negative, hence acceleration 9a) is negative, The speed (u) is positive in this interval dueto obvious reasons . ltbgt In interval ``, the velocity-time graph is parallelto time axis, therefore accelration 9a0 zero in this interval but (v) is positive due to obvious reasions.
AT points ` A , B` and C` the velocit-time graph is parallel to time axis . Therefore accelration (a0 iszero at all the four points.