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November 9th, 2016, 09:26 PM   #1
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Linear Velocity & Acceleration Question

I'm having trouble with this question.

A car traveling at a speed of 140km/h decelerates for 20 seconds. During this time it travels 500m.

a) What is its deceleration?

b) What is its speed after the deceleration?

Answers in the book are a) 1.389 m/s^2, b) 40 km/h

But I can't figure it out.
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November 9th, 2016, 10:02 PM   #2
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$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2$

You were given ...

$\Delta x = 500 \, m$, $t=20 \, s$, and $v_0=140 \, km/hr$

You need to convert km/hr to m/s and solve for acceleration, which will be a negative value.

Once you determine acceleration, you also have time and initial velocity.

use the equation ...

$v_f=v_0+at$

Final velocity will be in m/s ... you have to convert it to km/hr.
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November 23rd, 2016, 02:54 AM   #3
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Quote:
Originally Posted by skeeter View Post
$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2$

You were given ...

$\Delta x = 500 \, m$, $t=20 \, s$, and $v_0=140 \, km/hr$

You need to convert km/hr to m/s and solve for acceleration, which will be a negative value.

Once you determine acceleration, you also have time and initial velocity.

use the equation ...

$v_f=v_0+at$

Final velocity will be in m/s ... you have to convert it to km/hr.
My m/s I got 140 x 1000 / 3600 to get m/s = 38.89
38.89 / 20 seconds = 1.94 m/s as my Deceleration.... which is off my a bit.

Second questions what do you mean by at?

Displacement / Time = Velocity so. 500 / 20 = 25 m/s Then I divide by 1000 to convert to km's as well as x by 3600

500 / 20 = 25
25 / 1000 = .025
.025 x 3600 = 90 km's

So I'm off there as well

Last edited by jayinwww; November 23rd, 2016 at 02:59 AM.
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November 23rd, 2016, 05:05 AM   #4
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You didn't use the equations I cited in my previous post ... fyi, $a \ne \dfrac{v_0}{t}$, it is $a=\dfrac{v_f-v_0}{t}$. You can't use this equation to calculate acceleration because you do not have final velocity.

$140 \, km/hr \cdot \dfrac{1000 \, m}{km} \cdot \dfrac{1 \, hr}{3600 \, sec} = \dfrac{350}{9} \, m/s$

$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2 \implies a= \dfrac{2(\Delta x - v_0 \cdot t)}{t^2}$

$a= \dfrac{2 \left(500 - \frac{350}{9} \cdot 20 \right) }{20^2} = -\dfrac{25}{18} \, m/s^2$

Note $-\dfrac{25}{18} \approx -1.389$


displacement/time is average velocity, not final velocity ...

$v_f= v_0 + at$ ... this equation is a direct result of solving the equation $a=\dfrac{v_f-v_0}{t}$ for $v_f$

$v_f = \dfrac{350}{9} + \left(-\dfrac{25}{18}\right) \cdot 20 = \dfrac{100}{9} \, m/s$

$\dfrac{100}{9} \, m/s \cdot \dfrac{3600 \, sec}{1 \, hr} \cdot \dfrac{1 \, km}{1000 \, m} = 40 \, km/hr$

Final note ... if you are studying kinematics under uniform acceleration, you should have been exposed to these equations and the hints on when to use them.

Thanks from greg1313 and jayinwww

Last edited by skeeter; November 23rd, 2016 at 05:12 AM.
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