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May 11th, 2015, 06:26 AM   #1
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Constant Acceleration

A racing car is travelling at $130$ mph when the driver sees a broken-down car on the track $\frac{1}{10}$ of a mile ahead. Slamming the brakes on he achieves his maximum deceleration of $24.5$ mph per second. How far short of the broken-down car does he stop?

My attempt,
I know $u=130$, $a=-24.5$, which formula should I use to solve this question?
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May 11th, 2015, 04:07 PM   #2
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Quote:
Originally Posted by jiasyuen View Post
A racing car is travelling at $130$ mph when the driver sees a broken-down car on the track $\frac{1}{10}$ of a mile ahead. Slamming the brakes on he achieves his maximum deceleration of $24.5$ mph per second. How far short of the broken-down car does he stop?

My attempt,
I know $u=130$, $a=-24.5$, which formula should I use to solve this question?
.
First convert a to use one unit for time (hour - using mixed makes things complicated).

Get time (t) from u-at=0. then get distance (s) from $\displaystyle s=ut-\frac{at^2}{2}$.
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May 12th, 2015, 04:53 AM   #3
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I can't get the answer after figuring very long time. The answer is 0.0042 miles.

How to convert 24.5 mph per second to 25.5 miles/h^2?

Last edited by jiasyuen; May 12th, 2015 at 05:04 AM.
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May 12th, 2015, 05:20 AM   #4
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Using mixed units can be confusing, but I think this one isn't too bad.

It's a good rule of thumb to convert to SI when you begin. If you find that you've already performed a bunch of calculations and you don't want to redo everything, then go back over your calculations and check the units at each step. Below is an example where the units have been considered at every calculation step.

initial speed = u = 130 mph
final speed = v = 0 mph

acceleration = a = -24.5 mph /s

Let's get the time taken to stop:

SUVAT equation:

$\displaystyle v = u+at$
$\displaystyle t = \frac{v-u}{a} = \frac{0-130}{-24.5} = \frac{130}{24.5} = 5.306$ s

This is in seconds because we have $\displaystyle [t] = \frac{mph}{(mph/s)}=\frac{mph \cdot s}{mph} = s$

Now we know the time taken to stop, we can get the distance travelled by the car as it's stopping:

SUVAT equation:

$\displaystyle s = ut + \frac{1}{2}at^2$
$\displaystyle = 130 \times 5.306 - \frac{24.5}{2}(5.306)^2$
$\displaystyle = 689.796 - 344.898$
$\displaystyle = 344.898$

but what are the units of this? We have

$\displaystyle [s] = mph \cdot s = \frac{miles \cdot s}{hr}$

miles seconds per hour? Lovely. Let's convert the hours into seconds. We know that there are 3600 seconds in an hour, so

$\displaystyle 1 \frac{miles \cdot s}{hr} = 1 \frac{miles \cdot s}{3600 s} = \frac{1}{3600} \frac{miles \cdot s}{s} = \frac{1}{3600} miles$

So

$\displaystyle = 344.898 mph \cdot s = \frac{344.898}{3600} miles = 0.0958 miles$

So the distance, $\displaystyle d$, between the broken down car and the speeding car is:

$\displaystyle d = 0.1 - 0.0958 = 0.0042 miles$
Thanks from jiasyuen

Last edited by Benit13; May 12th, 2015 at 05:23 AM.
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May 12th, 2015, 05:52 AM   #5
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Benit. Thanks for your explanation.
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May 12th, 2015, 06:31 PM   #6
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Quote:
Originally Posted by jiasyuen View Post
I can't get the answer after figuring very long time. The answer is 0.0042 miles.

How to convert 24.5 mph per second to 25.5 miles/h^2?
Multiply by sec/hr (3600).
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