# If an object with uniform acceleration (or deceleration) has a speed of 12 m/s at t=0 and moves a total of 8 m by t=1, what was the object's rate of acceleration?

Jul 2, 2017

${a}_{x} = - 8$ ${\text{m/s}}^{2}$

#### Explanation:

We're asked to find te magnitude of the acceleration of an object with some given kinemtics quantities.

We know:

• initial velocity ${v}_{0 x}$ is $12$ $\text{m/s}$

• change in position $\Delta x$ is $8$ $\text{m}$

• time $t$ is $1$ $\text{s}$

We can use the equation

$\Delta x = {v}_{0 x} t + \frac{1}{2} {a}_{x} {t}^{2}$

Plugging in known values, and solving for the acceleration ${a}_{x}$, we have

8color(white)(l)"m" = (12color(white)(l)"m/s")(1color(white)(l)"s") + 1/2(a_x)(1color(white)(l)"s")^2

8color(white)(l)"m" = (12color(white)(l)"m") + 1/2(a_x)(1color(white)(l)"s")^2

-4color(white)(l)"m" = 1/2(a_x)(1color(white)(l)"s")^2

-8color(white)(l)"m" = (a_x)(1color(white)(l)"s"^2)

a_x = color(blue)(-8 color(blue)("m/s"^2

The acceleration is in the negative direction, which means the object will keep slowing down in the positive direction until it eventually slows to a stop, and it then speeds up in the negative direction.