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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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\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

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