You are standing at rest at a bus stop. a bus moving at a constant speed of 5.00 m/s passes you. when the rear of the bus is 11.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m/s2. a) how far would you have to run before you catch up with the rear of the bus? b) how fast must you be running then? find the final speed just as you reach the bus. c) would an average college student be physically able to accomplish this? yes or no

A) You would have to run 72.6 m.

B) You would have to run at a final speed of 11.8 m/s or 42.5 km/h.

C) No.

Explanation:

Hi there!

The equations for an object moving in a straight line with constant acceleration are as follows:

x = x0 + v0 * t + 1/2 a * t²

v = v0 + a * t

where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the movement is with constant velocity, the same equations apply only that a = 0.

A) You will catch the bus when your position is the same as the position of the bus.

The equation of your position will be:

x = x0 + v0 * t + 1/2 a * t²    

Since the center of the reference system is located at the point where you start running, x0 = 0. The initial velocity is 0 because you start from rest. Then:

x = 1/2 a * t²

The position of the bus will be:

x = x0 + v * t

Since you start running when the bus is 11.0 m past you, the initial position of the bus is 11.0 m.

If you catch the bus, then:

your position = position of the bus

1/2 a * t² = x0 + v * t

1/2 * 0.960 m/s² * t² = 11.0 m + 5.00 m/s * t

0.48 m/s² * t² - 5.00 m/s * t - 11.0 m

Solving the quadratic equation:

t = 12.3 s

Using t = 12.3 in the equation of your position, we will obtain the distance you have to run to catch the bus:

x = 1/2 * 0.960 m/s² * (12.3 s)² = 72.6 m

B) The equation for velocity is:

v = v0 + a * t

Since you start from rest, v0 = 0. Then, the velocity when you reach the bus at t = 12.3 s will be:

v = a * t

v = 0.960 m/s² * 12.3 s = 11.8 m/s

C) The velocity you have to run is almost the world record of Usain Bolt (45 km / h).

11.8 m/s *(3600 s/ h) *( 1 km / 1000 m) = 42.5 km/h.

Then, no, an average college student would not be able to run that fast.

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explanation:

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