Question

(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 2.57.

(b) Identify the time or times ( ta, tb, tc , etc.) at which the acceleration is greatest.

(c) At which times is it zero?

(d) At which times is it negative?

Short answer

a) The acceleration can be determined using the velocity versus time graph.

b) The acceleration is the greatest at b.

c) The acceleration is zero at points d, e, and h.

d) The acceleration is negative at the points I, j, k, l.

Step-by-step solution

Determination of acceleration

Acceleration is the rate of change in the velocity of a body with respect to time.

The slope of a graph of velocity vvs. time tis acceleration a.

$\mathbf{a}\mathbf{=}\frac{\mathbf{\Delta V}}{\mathbf{\Delta T}}\mathbf{=}\mathbf{slope}$

Hence, in the above graph, we need to find out the slope at each points from a to l .

A slope is an angle made by a tangent at a point to that of the angle made with the x-axis.

Hence, if we draw the tangent for all the points, we will get the figure below.

(a)

Acceleration at different points

Here, we can see that the slope becomes zero when it is at d, e, and h. Hence, at this point, there is no change in velocity, so acceleration will be zero at this point. From point a till d, the value of the slope gradually decreases and reaches zero, then the slope goes to the negative value from point d to e. After which the slope gradually increases till point h from point e. Once it reaches h, the slope again tends to reduce to the negative. Hence, the acceleration vs. time graph will be similar to the one shown below.

Figure: Acceleration VS time graph

Thus, we can conclude:

b) From the graph, it is clear that at b, the acceleration is the maximum.

c) The acceleration is zero at points d, e, and h.

d) The acceleration is negative at the points i, j, k, and l.