# If an object with uniform acceleration (or deceleration) has a speed of 5 m/s at t=0 and moves a total of 125 m by t=8, what was the object's rate of acceleration?

Jan 28, 2016

The acceleration is $\text{1 m/s"^2}$.

#### Explanation:

We will use the following equation to determine acceleration.

$\textcolor{red}{a = \frac{{v}_{f} - {v}_{i}}{\Delta t}}$,

where $a = \text{acceleration}$, ${v}_{f} = \text{final velocity}$, ${v}_{i} = \text{initial velocity}$, and $\Delta t = \text{time interval}$.

However, we must first determine the final velocity using the following equation.

$\textcolor{b l u e}{{v}_{f} = \frac{\Delta d}{\Delta t}}$,

where $\Delta d = \text{change in position}$, and $\Delta t = \text{time interval}$.

${v}_{f} = \left(125 \text{m")/(8"s}\right)$

${v}_{f} = \text{15.625 m/s}$

Now that we have the final velocity, we can use the first equation to determine the acceleration.

$\textcolor{red}{a = \frac{{v}_{f} - {v}_{i}}{\Delta t}}$

color(red)(a=((15.625"m/s")-(5"m/s"))/(8"s"))

$\textcolor{red}{a = \left(10.625 \text{m/s")/(8"s}\right)}$

$\textcolor{red}{a = \text{1 m/s"^2}}$ (rounded to one significant figure)