A car accelerates at 2.25 m/s^2 along a straight road. It passes two marks that are 27.8 m apart at times t=3.80 s and t=5.25 s. What was the car's velocity at t=0?

Answer 1

Let distances, S1 and S2, correspond to the distances at t1=3.8 and t2=5.25 respectively, assuming the distance =0 at t=0.

acceleration, a = 2.25 m/s/s
S1=u*(t1)+(1/2)a(t1)²
S2=u*(t2)+(1/2)a(t2)²

S2-S1=u(t2-t1)+(1/2)a(t2²-t1²)
Since S2-S1, t1,t2 and a are known, u can be isolated and solved for.