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April 10th, 2015, 05:38 AM   #1
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Kinematics and Newton's Laws

Looking for feedback in the quest to really understand kinematics

A train with mass $\displaystyle 3.6\times{10^5}kg$ travels with a velocity of $\displaystyle 120km/h = 33\frac{1}{3}m/s$.
Superman needs to stop the train in $\displaystyle 150m$.
How much force does Superman need to exert?

1) Calculate train's decelration

$\displaystyle v_x^2 = v_{x_0}^2 + 2a_x(x-x_0)$

$\displaystyle 0^2 = (33\frac{1}{3})^2 + 2a_x(150)$

$\displaystyle -1111\frac{1}{9} = 300a_x$

$\displaystyle a_x = -3\frac{19}{27}m/s^2$

2)Newton's Law

As a result of Superman pushing against the train, the train decelerates constantly at $\displaystyle -3\frac{19}{27}m/s^2$. Superman is therefore accelerating against the train at $\displaystyle 3\frac{19}{27}m/s^2$.

$\displaystyle F_{superman} = m_{train}\cdot{a}$

$\displaystyle F_{superman} = 3.6\times{10^5}\cdot{3\frac{19}{27}}$

$\displaystyle F_{superman} = \frac{4\times{10^6}}{3}N$ $\displaystyle \leftarrow$This is my answer after a second attempt.

For my first attempt I miscalculated the trains deceleration:

1)Time taken to decelerate to $\displaystyle 0m/s$,

$\displaystyle \bigtriangleup{v} = \frac{\bigtriangleup{x}}{\bigtriangleup{t}}$

$\displaystyle \bigtriangleup{t} = \frac{\bigtriangleup{x}}{\bigtriangleup{v}}$

$\displaystyle t = \frac{0-150}{0-33\frac{1}{3}}$

$\displaystyle t = 4.5s$

2)Trains deceleration

$\displaystyle a = \frac{\bigtriangleup{v}}{\bigtriangleup{t}}$

$\displaystyle a = \frac{0-33\frac{1}{3}}{4.5}$

$\displaystyle a = \frac{200}{27}m/s^2 = 7.407m/s^2$

$\displaystyle F_{superman} = \frac{8\times{10^6}}{3}N$

In this instance, $\displaystyle t=4.5s$ to slow from $\displaystyle 33\frac{1}{3}$ to $\displaystyle 0$ is incorrect.
The correct value is $\displaystyle t=9.0s$.

Why does my first attempt give an incorrect answer??

Thanks in advance

Last edited by hyperbola; April 10th, 2015 at 06:24 AM.
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April 10th, 2015, 06:57 AM   #2
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The use of

$\displaystyle a = \frac{\Delta v}{\Delta t}$

or

$\displaystyle v = \frac{\Delta x}{\Delta t}$

only gives exact answers when the quantities on the right-hand side vary linearly with time. In all other instances they only approximate the true result, with the approximation improving if the quantity varies slowly or is approximately linear.

SUVAT equations work for any situation where the acceleration is constant with time. This is true in your situation because Superman is providing a constant force and the train has constant mass, so due to Newton's second law the acceleration will be constant.

If the acceleration is constant, the velocity will be linear, but the displacement ($\displaystyle \Delta x$) will then change in a non-linear manner. You can convince yourself of this by calculating the position of the train at t = 0, 1, 2, 3 and 4 seconds. You'll find that the distance travelled each second will not be constant.

This is why your second attempt works but your first one doesn't. The fact your answer is exactly a factor of 2 out from the actual answer is coincidental.
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April 10th, 2015, 07:02 AM   #3
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$\displaystyle v_{avg.} = \frac{\bigtriangleup{x}}{\bigtriangleup{t}}$

$\displaystyle 33\frac{1}{3} = \frac{150}{t}$

This is incorrect because the initial velocity $\displaystyle (33\frac{1}{3})$ does not equal average velocity?

$\displaystyle v_{avg.} = \frac{v+v_0}{2}$

$\displaystyle v_{avg.} = \frac{33\frac{1}{3}}{2}$

$\displaystyle v_{avg.} = 16\frac{2}{3}m/s$

Therefore

$\displaystyle v_{avg.} = \frac{\bigtriangleup{x}}{\bigtriangleup{t}} $

$\displaystyle 16\frac{2}{3} = \frac{150}{t}$

$\displaystyle t = \frac{150}{16\frac{2}{3}}$

$\displaystyle t = 9.0s$

Correct in this instance??

Last edited by hyperbola; April 10th, 2015 at 07:08 AM.
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April 10th, 2015, 07:05 AM   #4
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Hey Benit13,

You posted whilst I was typing up my correction.

I'm going to read your post now.

Thanks again for your reply

Last edited by hyperbola; April 10th, 2015 at 07:08 AM.
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April 10th, 2015, 07:56 AM   #5
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No worries.

The correction you typed is wrong as well because it is coincidental that you got the factor of 2. Just don't use

$\displaystyle v = \frac{\Delta x}{\Delta t}$

at all in a SUVAT problem. It is useful to estimate average speed over some time interval, but you should use SUVAT equations to calculate anything exactly.

The equation above only works exactly when acceleration is zero so that the velocity is constant and the displacement will vary linearly with time. However, in this case it is basically just the same thing as

$\displaystyle speed = \frac{distance}{time}$
Thanks from hyperbola

Last edited by Benit13; April 10th, 2015 at 07:58 AM.
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April 10th, 2015, 08:46 AM   #6
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Okay. Thanks Benit13.

Main points taken on board.

I've only just started with calculus so just trying to assess my understanding.

CORRECT SOLUTION.
With reference to the train in my initial post, and Benit13's initial post:



Graph 1
accelerations vs. time
acceleration is constant

$\displaystyle \int a\cdot{dt} = velocity = Graph 2$

Graph 2
velocity vs. time
velocity decreases at a constant rate

$\displaystyle \int v\cdot{dt} = displacement = Graph 3$

Graph 3
displacement vs. time
non-linear

$\displaystyle x(t)=x_0 + v_0t + \frac{at^2}{2}$

$\displaystyle x(0) = 0$

$\displaystyle x(1) = 34\frac{23}{27}$

$\displaystyle x(2) = 74\frac{2}{27}$

$\displaystyle x(3) = 116\frac{2}{3}$

$\displaystyle x(4) = 162\frac{26}{27}$


INCORRECT SOLUTION



Graph 1
acceleration vs. time
acceleration is not constant

$\displaystyle a=\frac{\bigtriangleup{v}}{\bigtriangleup{t}}$

$\displaystyle \int \frac{\bigtriangleup{v}}{\bigtriangleup{t}}.dt = velocity = Graph 2$


Graph 2
velocity vs. time
non-linear
Attached Images
File Type: jpg graph.jpg (10.1 KB, 9 views)
File Type: jpg graph1.jpg (7.4 KB, 9 views)

Last edited by hyperbola; April 10th, 2015 at 09:06 AM.
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