Question

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.3 seconds, coasts for 3.0 s , and then slows down at a rate of 1.5m/s2 for the next stop sign.

a) How far apart are the stop signs?

Express your answer with the appropriate units.

Answer 1

during acceleration

the car starts from rest hence, initial velocity (u)= 0 m/s

acceleration (a) = 2.0 m/s²

time (t) = 6.3 s

distance traveled (s₁) = ut + 1/2 at²

or, s₁ = 0 x 6.3 + 1/2 x 2.0 x (6.3)² = 0 + 1 x (6.3)²

or, s₁ = 0 + 39.69

or, s₁ = 39.69 m

and final speed (v) = u + at = 0 + 2 x 6.3

or, v = 12.6 m/s

during motion with constant speed (coast)

time (t) = 3.0 s

now, the speed (v) = 12.6 m/s

distance covered (s₂) = v x t = 12.6 x 3

or, s₂ = 37.8 m

during retardation

a = -1.5 m/s²

initial velocity (u) = 12.6 m/s

since, car is going to be stopped hence, final velocity (v) = 0 m/s  

let , s₃ be the distance covered in this period

hence, v² - u² = 2 a s₃

or, 0² - (12.6)² = 2 x (-1.5) x s₃

or, 0 - 158.76 = - 3.0 x s₃

or, s₃ =  158.76 / 3.0

or, s₃ = 52.92 m

hence, total distance covered in entire journey (s) = s₁ + s₂ + s₃

or, s = 39.69 + 37.8 + 52.92

or, s = 130.41 m

hence, stop signs are 130.41 meter apart .