Question

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 21 m. The car passes the first sign at t = 1.3 s, the second sign at t = 3.3 s, and the third sign at t = 5.0 s.

(c) What is the magnitude of the acceleration of the car?

Answer 1

so consider first sign to be the zero

then at second sign, t = 2 s and d = 21 m

so x = x0 + v0 t + 1/2 a t^2

21 = v*2 + 0.5*a*2^2

v = 10.5 - 0.5*a*2

at third sign, t = 5-1.3 = 3.7 s, x = 2*21 = 42

42 = v*3.7 + 0.5*a*3.7^2

42 = (10.5 - 0.5*a*2)*3.7 + 0.5*a*3.7^2

a=1 m/s^2