Answer:
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.
