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

1 Answer
Apr 16, 2016

0.03ms2

Explanation:

Recall the following kinematic formula of motion:

∣ ∣ ∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aaΔd=viΔt+12aΔt2aa−−−−−−−−−−−−−−−−−−−−−

where:
Δd=change in distance
vi=initial velocity
Δt=change in time
a=acceleration

Start by listing out the given and required values.

vi=3ms

Δd=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.

Δd=viΔt+12aΔt2

=12aΔt2=ΔdviΔt

=2(ΔdviΔt)Δt2

=2(25m(3ms(8s)))(8s)2

=0.03125ms2

∣ ∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aa0.03ms2aa−−−−−−−−−