College Physics by Openstax Chapter 2 Problem 62

By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s2 at t=10 s .

Figure 2.63

Solution:

To solve for the slope of the curve at t=10 s, we need two points — 1 just before and 1 just after.

When t=0 st=0 \ \text{s}, v=166 m/sv=166 \ \text{m/s} and when t=20 st=20 \ \text{s}, v=230 m/sv=230 \ \text{m/s}. Therefore, the acceleration is

acceleration=slopeacceleration=ΔvΔta=v2−v1t2−t1a=230 m/s−166 m/s20 s−0 sa=3.2 m/s2\begin{align*} \text{acceleration} & =\text{slope} \\ \text{acceleration} & =\frac{\Delta v}{\Delta t} \\ \text{a} & = \frac{v_2-v_1}{t_2-t_1} \\ \text{a} & = \frac{230\ \text{m/s}-166\ \text{m/s}}{20\ \text{s}-0\ \text{s}} \\ \text{a} & =3.2\ \text{m/s}^2 \\ \end{align*}

Note that the values are approximated to satisfy the given acceleration in the problem statement. The values may differ from one’s answer due to some uncertainties of a graph.

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