College Physics by Openstax Chapter 2 Problem 18

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

a‾=ΔvΔt\overline{a}=\frac{\Delta v}{\Delta t}

If we rearrange the formula by solving for Δt\Delta t, in terms of velocity and acceleration, we come up with

Δt=Δva‾\Delta t=\frac{\Delta v}{\overline{a}}

Substituting the given values, we have

Δt=Δva‾Δt=vf−v0a‾Δt=2.00 m/s−0 m/s1.40 m/s2Δt=1.43 seconds  (Answer)\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

a‾=ΔvΔt\overline{a}=\frac{\Delta v}{\Delta t}

Then substituting all the given values, we have

a‾=vf−v0Δta‾=0 m/s−2 m/s0.8 m/s2a‾=−2.50 m/s2  (Answer)\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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