A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its velocity is 12.0 m/s. What is the distance traveled in that interval of time?

It begins at 8 m/s when t =5 seconds and ends at 12 m/s when t = 8 seconds

so

V0=8m/s

V1=12m/s

t0=5s

t1=8s

S - displacement or distance

S=((V0+V1)/2)*(t1-t0)

S = (( 8 + 12)/2)*(8-5) m

S = (10*3 )m

S = 30 m

I will be submitting to you an alternative approach using acceleration, in the following discussion

t = 8 - 5 = 3 s

a = 4/3 m/s2

S = ut + 1/2at^2

S = 8(3) + 1/2 x 4/3 x 3 x 3

= 24 + 6

= 30 m.

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