Figure 2.15 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 ?

Question

A particle moves in a circle of radius 2 m with a speed of 6 m/s. What is the centripetal acceleration of the particle?

Answer 1

1. Interval with greatest average acceleration (in magnitude)

Average acceleration over an interval = change in speed / time = slope of the speed–time graph over that interval.
From the graph, the slope is steepest (largest magnitude) in interval 2 (between B and C).

The average acceleration is greatest in magnitude in interval 2.

Average speed over an interval = (area under speed–time curve in that interval) / (time of interval).
In the graph, speeds are highest for most of the time in interval 3, so its average speed is largest.

The average speed is greatest in interval 3.

Speed is always positive, so velocity \(v\) (along the chosen positive direction) is positive in all three intervals.
Acceleration sign is given by slope of the speed–time graph:
- Interval 1: speed increases ⇒ slope > 0 ⇒ \(a > 0\).
- Interval 2: speed decreases ⇒ slope < 0 ⇒ \(a < 0\).
- Interval 3: speed is constant ⇒ slope = 0 ⇒ \(a = 0\).

Instantaneous acceleration at a point = slope of the tangent to the speed–time curve at that point.
At A, B, C, and D, the curve has “corners” where the graph changes segment; the tangent is not well-defined.
In the NCERT-style treatment, these points are treated as transition points, and the acceleration is taken as zero there (slope changes abruptly, but over an instant the change in speed is taken as zero).

Accelerations at A, B, C and D are taken as zero.