College Physics by Openstax Chapter 2 Problem 16

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

Solution:

The formula for acceleration is 

a‾=change in velocitychange in timea‾=ΔvΔta‾=vf−v0tf−t0\begin{align*} \overline{a} & =\frac{\text{change in velocity}}{\text{change in time}}\\ \overline{a} & =\frac{\Delta \text{v}}{\Delta t} \\ \overline{a} & =\frac{v_f-v_0}{t_f-t_0} \\ \end{align*}

Substituting the given values

a‾=30.0 m/s−0 m/s7.00 s=4.29 m/s2  (Answer)\begin{align*} \overline{a} & =\frac{30.0\:\text{m/s}-0\:\text{m/s}}{7.00\:\text{s}} \\ & =4.29\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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