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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.
A. For what time interval (in s) after the light turned green is the bicycle ahead of your car?
B. By what maximum distance does the bicycle lead the car?

Sagot :

oladayoademosu

Answer:

A. 2.63 s B. 12.38 m

Explanation:

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

The time interval at which the bicycle is ahead of the car is the time it takes for the car to reach the bicycle's speed of 21.0 mi/h.

So, using v = u + at where u = initial speed of car = 0 mi/h, v = final speed of car = 21.0 mi/h, a = acceleration of car = 8.00 mi/h/s and t = time taken for acceleration.

So, v = u + at

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (21.0 mi/h - 0 mi/h)/8.00 mi/h/s

= 21.0 mi/h ÷ 8.00 mi/h/s

= 2.63 s

B. By what maximum distance does the bicycle lead the car?

To find this distance, we find the distance moved by both the car in this time of t = 2.63 s

So, using s = ut + 1/2at² where u = initial speed of car = 0 mi/h = 0 m/s, t = time = 2.63 s, a = acceleration of car = 8.00 mi/h/s = 8.00 × 1609 m/3600 s = 3.58 m/s/s = 3.58 m/s² and s = distance moved by car.

So, substituting the values of the variables into the equation, we have

s = ut + 1/2at²

s = 0 m/s × 2.63 s + 1/2 × 3.58 m/s² × (2.63 s)²

s = 0 m + 1/2 × 3.58 m/s² × (2.63 s)²

s =  1.79 m/s² × 6.9169 s²

s = 12.38 m

which is also the maximum distance with which the bicycle leads the car.

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