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?

1 Answer

Answer:

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#