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 3.4 m/s² for 4.1 seconds. It then continues at a constant speed for 14.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 258.93 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 2.5 seconds after it starts?m/s

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2)

How fast is the blue car going 9 seconds after it starts?m/s

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3)

How far does the blue car travel before its brakes are applied to slow down? m

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4)

What is the acceleration of the blue car once the brakes are applied? m/s²

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5)

What is the total time the blue car is moving?s

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6)

What is the acceleration of the yellow car?

Answer 1

1)
Initial velocity, u = 0
Acceleration, a = 3.4 m/s²
Time taken, t = 2.5 s
Consider v as the velocity at t = 2.5 s

Using the formula, v = u + at,
v = 0 + 3.4 * 2.5
= 8.5 m/s

2)
We need to find the velocity at t = 4.1 s since the blue car was going with the same velocity from 4.1 s to 14.7 s.

Initial velocity, u = 0
Acceleration, a = 3.4 m/s²
Time taken, t = 4.1 s
Consider v as the velocity at t = 4.1 s

Using the formula, v = u + at,
v = 0 + 3.4 * 4.1
= 13.94 m/s

3)
Distance travelled by the blue car in 4.1 s, s = u * t + 0.5 * a * t²
= 0 * 4.1 + 0.5 * 3.4 * 4.1²
= 28.577 m

Distance travelled by the blue car from 4.1 s to 14.7 s,
s' = v * (14.7 - 4.1)
Where v is the velocity of the blue car at 4.1 s
s' = 13.94 * (14.7 - 4.1)
= 147.764 m

Total distance travelled by the blue car before the brakes were applied,
d = s + s'
= 28.577 + 147.764
= 176.34 m

4)
Distance travelled by the blue car during deceleration, d' = D - d
Where D is the total distance travelled by the blue car.
d' = 258.93 - 176.34
= 82.59 m
Initial velocity, v = 13.94 m/s
Final velocity, vf = 0
Consider a' as the acceleration

Using the formula, vf² - v² = 2 * a' * d'
0 - 13.94² = 2 * a' * 82.59
a' = - 1.18 m/s² (negative sign shows that the car is decelerating)

5)
Consider t' as the time taken by the blue car during the deceleration.
Initial velocity, v = 13.94 m/s
Final velocity, vf = 0
Acceleration, a' = - 1.18 m/s²
Consider t' as the time taken.

Using the formula, vf = v + a * t',
t' = (vf - v) / a
= (0 - 13.94) / (-1.18)
= 11.85 s

Total time the blue car was moving, T = 14.7 + t'
= 14.7 + 11.85
= 26.55 s

6)
Initial velocity of the yellow car, ui = 0
Time taken by the yellow car, T = 26.55 s
Distance travelled by the yellow car, D = 258.93 m

Using the formula, D = ui * t + 0.5 * a_y * T²,
Where ay is the acceleration of the yellow car.
258.93 = 0 + 0.5 * a_y * 26.55²
a_y = 0.735 m/s²