# If an object with uniform acceleration (or deceleration) has a speed of 3 m/s at t=0 and moves a total of 58 m by t=6, what was the object's rate of acceleration?

Feb 21, 2016

a=20/9 m/s^2
a=-20/9 m/s^2

#### Explanation:

$\text{area under graph is equal displacement of 58 m}$
$58 = \frac{3 + k}{2} \cdot 6 \text{; "58=(3+k)*3"; } 58 = 9 + 3 \cdot k$
$49 = 3 \cdot k \text{; } k = \frac{49}{3}$
$\text{rate of acceleration is equal tangent of graph}$
$a = \frac{3 - k}{6} \text{ or } a = \frac{k - 3}{a}$
$a = \frac{3 - \frac{49}{3}}{6} \text{ ; } a = - \frac{20}{9} \frac{m}{s} ^ 2$
$a = \frac{\frac{49}{3} - 3}{6} \text{ ; } a = \frac{20}{9} \frac{m}{s} ^ 2$