(a) The car decelerated at an acceleration of -3 m/s^2.
(b) If maintaining the same deceleration, the car would stop before reaching the lights.
(c) It took approximately 2.67 seconds before the lights changed to green.
(a) Acceleration while slowing down:
The initial velocity (u) is 20 m/s, the final velocity (v) is 8 m/s, and the displacement (s) is the distance covered from 200 m to 50 m. The acceleration (a) can be found using the equation v^2 = u^2 + 2as.
a = (v^2 - u^2) / (2s) = (8^2 - 20^2) / (2 * (50 - 200)) = -3 m/s^2
The negative sign indicates deceleration.
(b) Maintaining the same acceleration:
If the car maintains the same deceleration, it will eventually come to a stop. Since the final velocity (v) will be zero, the equation v^2 = u^2 + 2as can be used to find the stopping distance (s).
0 = 20^2 + 2 * (-3) * s
s = 20^2 / (2 * 3) = 66.67 m
The car would stop before reaching the lights.
(c) Time before lights changed to green:
The car covers the distance from 50 m to 0 m (lights) under acceleration. The equation v^2 = u^2 + 2as can be used again to find the time (t) taken.
15^2 = 8^2 + 2 * a * s
a = (15^2 - 8^2) / (2 * s)
Substituting this a into the kinematic equation v = u + at, where v = 15 m/s (final velocity), u = 8 m/s (initial velocity), and s = 50 m (displacement):
15 = 8 + ((15^2 - 8^2) / (2 * s)) * t
Solving for t, we find t ≈ 2.67 seconds.
The question probable may be
A car travelling at 20 m s-¹ was 200 m from a set of traffic lights, which were red. The driver slows down until he was 50 m from the lights and travelling at 8 m s¹, when the lights changed to green. He immediately began to accelerate, passing the lights at a speed of 15 m s-¹. (a) What was the acceleration of the car while it was slowing down? (b) If the driver had maintained the same acceleration would he have stopped before he reached the lights? (c) How long was it before the lights changed to green?