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Olin [163]
3 years ago
8

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?
Physics
1 answer:
Gennadij [26K]3 years ago
8 0

Answer:

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

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1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

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Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

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Toward us

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<em></em>

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