Two cars start from rest at a red stop light. When the light turns green, both c
ID: 1354192 • Letter: T
Question
Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 4.8 m/s2 for 4 seconds. It then continues at a constant speed for 9.9 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 260.35 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.
1) How fast is the blue car going 1.2 seconds after it starts?
2) How fast is the blue car going 6.8 seconds after it starts?
3) How far does the blue car travel before its brakes are applied to slow down?
4) What is the acceleration of the blue car once the brakes are applied?
5) What is the total time the blue car is moving?
6) What is the acceleration of the yellow car?
Explanation / Answer
(1)
As the blue car is accelerating for 4 seconds, we can use v = u + a * t
u = 0
a = 4.8
t = 1.2
=> v = 0 + 4.8 * 1.2 = 5.76 m/sec
At 1.2 seconds the blue car's velocity is 5.76 m/sec
(2)
As, the blue car is accelerating upto 4 seconds. Let's find it final velocity at 4 seconds
v = 0 + 4.8 * 4 = 19.2 m/sec
So, now the car will continue with the above uniform speed for next 9.9 seconds
=> Blue car's velocity at 6.8 seconds will be 19.2 m/sec
(3)
Total distance travelled before the brakes applied = distance travelled while accelerating + distance travelled with the constant velocity
Total distance = (u1 * t1 + 0.5 * a1 * t12) + (u2 * t2 + 0.5 * a2 * t22)
Distance = (0 + 0.5 * 4.8 * 4 * 4) + (19.2 * 9.9 + 0)
Distance = 228.48 meters
The distance travelled by the blue car before its brakes are applied = 228.48 meters
(4)
Given the total distance travelled by the blue car = 260.35 meters
Distance travelled while the brakes are applied = Total distance - Distance travelled before the brakes applied
Distance (s) = 260.35 - 228.48 = 31.87 meters
From v2 - u2 = 2 * a * s
=> a = v2 - u2 / 2 * s
=> a = 0 - (19.2)2 / (2 * 31.87)
=> a = -5.78 m/s^2
The acceleration of the blue car once the brakes are applied = -5.78 m/s^2
(5)
Time taken by the blue car once the breakes applied can be found from
v = u + a * t
0 = 19.2 - 5.78 * t
=> t = 19.2 / 5.78 = 3.32 seconds
Total time the blue car is moving = 4 + 9.9 + 3.32 = 17.22 seconds
(6)
From s = u * t + 0.5 * a * t2
u for yellow car is 0
t is the same time that blue car took for the whole journey = 17.22 seconds
s for the whole journey = 260.35 meters
Substituting
260.35 = 0 + 0.5 * a * (17.22)2
=> a = 1.756 m/s^2
The acceleration of yellow car = 1.756 m/s^2
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