# If an object with uniform acceleration (or deceleration) has a speed of 3 m/s at t=0 and moves a total of 8 m by t=4, what was the object's rate of acceleration?

May 7, 2016

Deceleration of $- 0.25 \frac{m}{s} ^ 2$

#### Explanation:

At time ${t}_{i} = 0$ it had initial velocity of ${v}_{i} = 3 \frac{m}{s}$

At time ${t}_{f} = 4$ it had covered $8 m$

So

${v}_{f} = \frac{8}{4}$

${v}_{f} = 2 \frac{m}{s}$

Rate of acceleration is determined from

$a = \frac{{v}_{f} - {v}_{i}}{{t}_{f} - {t}_{i}}$

$a = \frac{2 - 3}{4 - 0}$

$a = - \frac{1}{4}$$\frac{m}{s} ^ 2$

$a = - 0.25 \frac{m}{s} ^ 2$

As $a$ is negative we take it as deceleration of $- 0.25 \frac{m}{s} ^ 2$

Cheers