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
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
Thus, rearrange the formula for
Δd=viΔt+12aΔt2
=12aΔt2=Δd−viΔt
=2(Δd−viΔt)Δt2
=2(25m−(3ms(8s)))(8s)2
=0.03125ms2
≈∣∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aa0.03ms2aa∣∣∣−−−−−−−−−−−