# If an object is moving at 0 m/s and accelerates to 60 m/s over 34/5 seconds, what was the object's rate of acceleration?

Jun 26, 2016

$\vec{a} \approx 9 \frac{m}{s} ^ 2 \left[\text{forward}\right]$

#### Explanation:

The acceleration of an object is given by the formula:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\vec{a} = \frac{\Delta \vec{v}}{\Delta t}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\underline{\text{where:}}$
$\vec{a} =$acceleration, in metres per second squared $\left(\frac{m}{s} ^ 2\right)$
$\Delta \vec{v} =$change in velocity, in metres per second $\left(\frac{m}{s}\right)$
$\Delta t =$change in time, in seconds $\left(s\right)$

Since you are given the change in velocity as well as the change in time, you can go straight to solving for the object's acceleration.

In this case, we will let the forward direction be positive.

$a = \frac{\Delta v}{\Delta t}$

$\textcolor{w h i t e}{\times x} = \frac{{v}_{f} - {v}_{i}}{\Delta t}$

Substituting in the values,

$\textcolor{w h i t e}{\times x} = \frac{60 \frac{m}{s} - 0 \frac{m}{s}}{\frac{34}{5} s}$

$\textcolor{w h i t e}{\times x} = 8.82 \frac{m}{s} ^ 2$

$\textcolor{w h i t e}{\times x} \approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{9 \frac{m}{s} ^ 2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$