# If an object with uniform acceleration (or deceleration) has a speed of 3 m/s at t=0 and moves a total of 25 m by t=8, what was the object's rate of acceleration?

Apr 16, 2016

$0.03 \frac{m}{s} ^ 2$

#### Explanation:

Recall the following kinematic formula of motion:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta d = {v}_{i} \Delta t + \frac{1}{2} a \Delta {t}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
$\Delta d =$change in distance
${v}_{i} =$initial velocity
$\Delta t =$change in time
$a =$acceleration

Start by listing out the given and required values.

${v}_{i} = 3 \frac{m}{s}$

$\Delta d = 25 m$

$t = 8 s$

a=?

Notice how the only variable to solve for is $a$.

Thus, rearrange the formula for $a$ and plug in your known values into the formula to find the acceleration of the object.

$\Delta d = {v}_{i} \Delta t + \frac{1}{2} a \Delta {t}^{2}$

$= \frac{1}{2} a \Delta {t}^{2} = \Delta d - {v}_{i} \Delta t$

$= \frac{2 \left(\Delta d - {v}_{i} \Delta t\right)}{\Delta {t}^{2}}$

$= \frac{2 \left(25 m - \left(3 \frac{m}{s} \left(8 s\right)\right)\right)}{8 s} ^ 2$

$= 0.03125 \frac{m}{s} ^ 2$

$\approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 0.03 \frac{m}{s} ^ 2 \textcolor{w h i t e}{\frac{a}{a}} |}}}$