# If an object with uniform acceleration (or deceleration) has a speed of 4 m/s at t=0 and moves a total of 18 m by t=5, what was the object's rate of acceleration?

Apr 30, 2016

$a = 0 , 16 \text{ "m/s^2" acceleration}$
$a = - 0 , 16 \text{ "m/s^2" deceleration}$
$\left(\text{acceleration")/("deceleration}\right) = - 1$

#### Explanation:

$\text{in figure-1:}$
$\text{area under graph is equal to displacement of object}$

$18 = \frac{\left(4 + 4 + \Delta v\right) \cdot 5}{2}$

$36 = 40 + 5 \Delta v \text{ "5 Delta v=36-40" "5 Delta v=4 " " Delta v=4/5=0,8" } \frac{m}{s}$

$\tan \alpha = a = \frac{\Delta v}{\Delta t}$

$a = \frac{0 , 8}{5}$

$a = 0 , 16 \text{ "m/s^2" acceleration}$

$\text{in figure-2}$

$18 = \frac{\left(4 + 4 - \Delta v\right) \cdot 5}{2}$

$36 = \left(8 - \Delta v\right) 5$

$36 = 40 - 5 \Delta v$

$5 v = 40 - 36$

$5 v = 4$
$v = \frac{4}{5} = 0 , 8 \text{ } \frac{m}{s}$

$\tan \beta = a = - \frac{0 , 8}{5}$

$a = - 0 , 16 \text{ "m/s^2" deceleration}$