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 vand a in the three intervals. What are the acclerations at the points A, B, C and D?

Explanation

(i) Average acceleration `= ("change in speed")/("time")`
Now for smaller intervals of time `v to t` graph, slope gives average acceleration. In second part, the slope is maximum. Hence acceleration is maximum (more than part 1 and part 3).
(ii) The average speed is maximum in `3^(rd)` part but a is zer` a =0 , v gt 0`
(iii) In first part, slope is positive, Hence, a and v both are positive `a gt 0 , v gt 0`
(iv) In second part the slope is negative, hence a is negative and speed is positive `a lt 0, v gt 0.`
(v) Speed is positive and a is zero in third part of graph.
(vi) Acceleration will be zero at points A,B,C, D in graph.