Q2-31E

Question

The graph in Fig. E2.31 shows the velocity of a motorcycle police officer plotted as a function of time. (a) Find the instantaneous acceleration at \(t = 3\,s,\,\,t = 7\,s{\rm{ and }}t = 11\,s\).

Step-by-Step Solution

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Answer

Hence, the required accelerations are

\({a_3} = 0{\rm{ m/}}{{\rm{s}}^2},\,\,\,{a_7} = 6.25{\rm{ m/}}{{\rm{s}}^2},\,\,{\rm{and }}\,{a_{11}} = - 11.25{\rm{ m/}}{{\rm{s}}^2}\).

1Step 1: Velocity and Acceleration

The measurement of change of displacement with respect to time is known as velocity, whereas that of velocity with respect to time is known as acceleration.

2Step 2: Examine the given graph (a)

The graph for velocity of a motorcycle as a function of time is:

Now, at \(t = 3\,s\), the velocity is constant.

Therefore, acceleration at \(t = 3\,s\) will be zero.

And, from \(t = 3\,s{\rm{ to }}t = 7\,s\), there is change in the velocity.

Therefore, acceleration here will be:

\({a_7} = \frac{{\Delta v}}{{\Delta t}} = \frac{{45 - 20}}{{7 - 3}} = \frac{{25}}{4} = 6.25{\rm{ m/}}{{\rm{s}}^2}\)

3Step 3: Find acceleration

Similarly, from \(t = 7\,s{\rm{ to }}t = 11\,s\), there is change in the velocity.

Therefore, acceleration here will be:

\({a_{11}} = \frac{{\Delta v}}{{\Delta t}} = \frac{{0 - 45}}{{11 - 7}} = \frac{{ - 45}}{4} = - 11.25{\rm{ m/}}{{\rm{s}}^2}\)

Hence, the required acceleration are