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 12.0 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/s 2 . (a) how far would you have to run before you catch up with the rear of the bus, and (b) how fast must you be running then? 2 (c) would an average college student be physical able to accomplish this?

a. 74,14 m

b. 11,93 m/s o 42,95 km/h

c. Nope. Not even on your wildest dreams.

Explanation:

So, in order to solve this situation, we must set common ground between me and the bus I plan to catch. First, we know that the bus and me are going to be on the same place at the same time if I manage to catch it. So, we choose one kinematic equation that allow us to stablish this common ground. In this case, we use the equation:

x_bus = x_0bus + v_0bus*t + 1/2 * a_bus * t^2

x_me = x_0me + v_0me*t + 1/2*a_me * t^2

We already stablished that the bus and I are going to be at the same place

x_bus = x_me, so we can match both of our equations:

x_0bus + v_0bus*t + 1/2*a_bus*t^2 = x_0me + v_0me*t + 1/2*a_me*t^2

We will fixate the reference point at my initial position, that means that x_0me = 0. Additionally, we know that the bus has constant velocity and that I started from rest. (a_bus = 0;v_0me = 0). So, we apply these conditions to our equation:

x_0bus + v_0bus*t = 1/2*a_me*t^2

What remains is a quadratic equation. We replace the values and put all the terms on one side and the equations remains like this:

1/2*0,960*t^2 - 5*t - 12 = 0

We can find the roots of the equation using the general formula of the quadratic equation:

t = (-b±√(b^2-4ac))/2a = (-(-5)±√(〖(-5)〗^2-4*(0,48)*(12)))/(2*0,48)

Solving the quadratic formula, we find two roots; t1=-2.0116 s, which is before the whole situation started and have no physical value, and t2=12.4282 s. so it would take me 12.4282 second to catch the rear of the bus.

Using the first equation we set for the bus, we have that:

x_bus = x_me = 12 m + 5m/s * (12,4282 s) = 74,14 m

This mean that I would have to run 74 m in 12.4282 seconds to catch the rear of the moving bus. To find how fast I would have to run, we use the following relation:

v_me = v_0me + a_me*t = 0 + 0,960 m/s^2 *12.4282 s = 11,93 m/s

The average human reaches a velocity of 10 km/h when they run. So there is no way I could catch that bus because:

11,93 m/s * (1 km)/(1000 m) * (3600 s)/(1 h) = 42,95 km/h

I would have to run at 42,95 km/h to do it

Let me know if there is something else I can help you with. Have a great day! :D

the correct choice is a .

do not post any more test or exam questions on

i believe the answer is a hope it and is correct