If an object with uniform acceleration (or deceleration) has a speed of 3 m/s at t=0 and moves a total of 25 m by t=8, what was the object's rate of acceleration?

1 Answer
Johnson Z.
Apr 16, 2016

0.03m/s^2

Explanation:

Recall the following kinematic formula of motion:

color(blue)(|bar(ul(color(white)(a/a)Deltad=v_iDeltat+1/2aDeltat^2color(white)(a/a)|)))

where:
Deltad=change in distance
v_i=initial velocity
Deltat=change in time
a=acceleration

Start by listing out the given and required values.

v_i=3m/s

Deltad=25m

t=8s

a=?

Notice how the only variable to solve for is a.

Thus, rearrange the formula for a and plug in your known values into the formula to find the acceleration of the object.

Deltad=v_iDeltat+1/2aDeltat^2

=1/2aDeltat^2=Deltad-v_iDeltat

=(2(Deltad-v_iDeltat))/(Deltat^2)

=(2(25m-(3m/s(8s))))/(8s)^2

=0.03125m/s^2

~~color(green)(|bar(ul(color(white)(a/a)0.03m/s^2color(white)(a/a)|)))

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