# If an object with uniform acceleration (or deceleration) has a speed of 4 m/s at t=0 and moves a total of 45 m by t=3, what was the object's rate of acceleration?

Dec 30, 2015

$a = \frac{22}{3} \frac{m}{s} ^ 2$

#### Explanation:

Data:-
Initial velocity$= {v}_{i} = 4 \frac{m}{s}$
Distance $= S = 45 m$
Time taken $= t = {t}_{2} - {t}_{1} = 3 - 0 = 3 s$
Acceleration=a=??
Sol:-
We know that:-
$S = {v}_{i} \cdot t + \frac{1}{2} a {t}^{2}$
$\implies 45 = 4 \cdot 3 + \frac{1}{2} a {3}^{2}$
$\implies 45 = 12 + \frac{9}{2} a$
$\implies \frac{9}{2} a = 33 \implies 9 a = 66 \implies a = \frac{66}{9} = \frac{22}{3} \frac{m}{s} ^ 2$
$\implies a = \frac{22}{3} \frac{m}{s} ^ 2$