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You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 49.0 mi/h with constant acceleration 8.00 mi/h/s and thereafter moves with a constant speed of 49.0 mi/h. At the same time, the cyclist speeds up from rest to 21.0 mi/h with constant acceleration 14.00 mi/h/s and thereafter moves with a constant speed of 21.0 mi/h.

Required:
For what time interval (in s) after the light turned green is the bicycle ahead of your car?

Sagot :

Answer:

t = 4.34 s

Explanation:

To get this, we need to calculate the time of each part for both vehicles.

In the case of the bicycle, we can calculate the time duration with it's acceleration:

a = Vf - Vo/t ------> t = Vf - Vo / a

But Vo = 0 so time:

t = 21 / 14 = 1.5 s

Doing the same with the car:

t = 49 / 8 = 6.125 s

With these values of time, we can calculate the distance covered by both vehicles during acceleration:

X = Vo*t + at²/2

X = at² / 2

With the bicycle:

X = 14 * (1.5)² / 2 = 15.75 mi

With the car:

X = 8 * (1.5)² / 2 =  9 mi

And now, we can also get the speed of the car, after the time of 1.5 s has passed.

V = 8 * 1.5 = 12 mi/h

Now we can actually write an equation with both data of the vehicles in function of the distance. We are going to say that "t" would be the time taken by both vehicles, to meet each other.

Distance covered by the bicycle = distance covered by car

Distance covered by the bicycle would be:

15.75 mi + 21t

Distance covered by car:

9 + 12t + (8t²/2)

Equalling both expressions:

15.75 + 21t = 9 + 12t + 4t²

4t² - 9t - 6.75 = 0

Solving for t, using quadratic expressions we have:

t = 9 ± √(9)² + 4*4*6.75 / 2*4

t = 9 ±√189 / 8

t = 9 ± 13.75 / 8

t₁ = 2.84 s

t₂ = -0.594 s

We are taking the positive time.

Then, the time where both vehicles meet, which is the same time interval when the bicycle is ahead of the car will be:

t = 2.84 + 1.5

t = 4.34 s

Hope this helps

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