A commuter backs her car out of her garage with an acceleration of 1.40 m/s2 .
(a) How long does it take her to reach a speed of 2.00 m/s?
(b) If she then brakes to a stop in 0.800 s, what is her deceleration?
Solution:
Part A
The formula for acceleration is
\overline{a}=\frac{\Delta v}{\Delta t}If we rearrange the formula by solving for \Delta t, in terms of velocity and acceleration, we come up with
\Delta t=\frac{\Delta v}{\overline{a}}Substituting the given values, we have
\begin{align*}
\Delta t & =\frac{\Delta v}{\overline{a}} \\
\Delta t & = \frac{v_f-v_0}{\overline{a}} \\
\Delta t & =\frac{2.00 \ \text{m/s}-0 \ \text{m/s}}{1.40 \ \text{m/s}^2} \\
\Delta t & =1.43 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
The formula for acceleration (deceleration) is
\overline{a}=\frac{\Delta v}{\Delta t}Then substituting all the given values, we have
\begin{align*}
\overline{a} & = \frac{v_f-v_0}{\Delta t} \\
\overline{a} & = \frac{0 \ \text{m/s}-2\ \text{m/s}}{0.8 \ \text{m/s}^2} \\
\overline{a} & = -2.50 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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