While sledding down a snowy hill Ed slowed down from 5 m/s to rest in a distance of 100 m. What was Ed's acceleration?

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While sledding down a snowy hill Ed slowed down from 5 m/s to rest in a distance of 100 m. What was Ed's acceleration?

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Answer 1 · 2016-01-02T09:27:24

Since you also have time as an unknown value, you need 2 equations that combine these values. By using the equations of speed and distance for deceleration, the answer is:

$a=0.125 m/s^2$

Explanation:

1st way

This is the simple elementary path. If you are new to motion, you want to go this path.

Provided that the acceleration is constant, we know that:

$u=u_0+a*t" " " "(1)$

$s=1/2*a*t^2-u*t" " " "(2)$

By solving $(1)$ for $t$ :

$0=5+a*t$

$a*t=-5$

$t=-5/a$

Then substituting in $(2)$ :

$100=1/2*a*t^2-0*t$

$100=1/2*a*t^2$

$100=1/2*a*(-5/a)^2$

$100=1/2*a*(-5)^2/a^2$

$100=1/2*25/a$

$a=25/(2*100)=0.125 m/s^2$

2nd way

This path is not for beginners, as it is the calculus path. All it provides is actual proof of the above equations. I am just posting in case you are interested in how it works.

Knowing that $a=(du)/dt$ we can transform by using chain rule through Leibniz's notation:

$a=(du)/dt=(du)/dt*(dx)/dx=(dx)/dt*(du)/dx$

Knowing that $u=(dx)/dt$ gives us:

$a=u*(du)/dx$

By integrating:

$a*dx=u*du$

$aint_0^100dx=int_5^0udu$

$a*[x]_0^100=[u^2/2]_5^0$

$a*(100-0)=(0^2/2-5^2/2)$

$a=5^2/(2*100)=25/(2*100)=1/(2*4)=0.125 m/s^2$

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While sledding down a snowy hill Ed slowed down from 5 m/s to rest in a distance of 100 m. What was Ed's acceleration?

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Explanation:

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