# If an object with uniform acceleration (or deceleration) has a speed of 9 m/s at t=0 and moves a total of 44 m by t=3, what was the object's rate of acceleration?

Sep 1, 2016

$3. \dot{7} m {s}^{-} 2$

#### Explanation:

Given distance moved $s = 44 m$ in time $3 s$, initial speed at $t = 0$ is $9 m {s}^{-} 1$. To find acceleration $a$.
We can use kinematic equation connecting above variables
$s = u t + \frac{1}{2} a {t}^{2}$
Plugin given values
$44 = 9 \times 3 + \frac{1}{2} a {\left(3\right)}^{2}$
$\implies \frac{9}{2} a = 44 - 27$
$\implies \frac{9}{2} a = 44 - 27$
$\implies a = 3. \dot{7} m {s}^{-} 2$