A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.2 seconds, coasts for 2.5 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. How far apart are the stop signs

the distance between the stop signs is 120.7 m.

Explanation:

The car moved in three stages;

(1) It accelerates from rest at 2.0 m/s² for 6.2 seconds

(2) it moved at a constant speed for 2.5 s

(3) it finally decelerate at the rate of 1.5m/s²

(1) The distance moved by the car during the first stage;

s₁ = ut + ¹/₂at²

s₁ = 0 + ¹/₂ (2)(6.2)²

s₁ = 38.44 m

(2) The distance moved by the car during the second stage;

calculate the constant speed of the car,

v = u + at

v = 0 + 2 x 6.2

v = 12.4 m/s

The distance moved by the  car as it coasts for 2.5s: s₂ = vt

s₂ = 12.4 x 2.5

s₂ = 31 m

(3) The distance moved by the car during the third stage;

When the car stops, the final velocity is zero.

v² = u² + 2as₃

a = -1.5 m/s², since the car slowed down or decelerated.

0 = 12.4² + (2 x - 1.5)s₃

0 = 153.76 - 3s₃

3s₃ = 153.76

s₃ = 153.76 / 3

s₃ = 51.253 m

The total distance moved by the car from the start to stop = s₁ + s₂ + s₃

d = 38.44 m + 31 m + 51.253 m

d = 120.7 m

Therefore, the distance between the stop signs is 120.7 m.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y