# Hard physics problem

CellCoree
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.

Question A.) What is the acceleration of the sled? ___ m/s^2

Question B.) What is the speed of the sled when it passes the 14.4-m point? ___ m/s

it's hard to start this problem because i don't know the initial time, can someone help me with this problem?

needhelpperson
CellCoree said:
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.

Question A.) What is the acceleration of the sled? ___ m/s^2

Question B.) What is the speed of the sled when it passes the 14.4-m point? ___ m/s

it's hard to start this problem because i don't know the initial time, can someone help me with this problem?

ok set up the equations as
using vi*2 + .5*a*2^2 = 25.6-14.4

vi*4 + .5*a*4^2 = 40 - 14.4

vi*6 + .5*a*6^2 = 57.6 - 14.4
just solve these three system of equations for a or vi

and you should have the answers

(i think you only need to solve 2 of them)

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thermodynamicaldude
Needpersonhelp's method works, largely cause in this case the acceleration is constant. Another method that will work is outlined below:

14.4 m

25.6 m 2 s later

40 m Another 2 s later

57.6 m Another 2 s later

Now, find the differences in the distances...and you should get something like:

11.2 m for first time interval

14.4 m for second time interval

17.6 m for third time interval

Now, realize that each time inteveral is 2 seconds. Divide each of these distances by 2 seconds to get average velocity for those time intervals. Now, find the differences between THESE average velocities...and divide by two seconds to get accerleration. Now with acceleration, you can find initial velocity.

needhelpperson
thermodynamicaldude said:
Needpersonhelp's method works, largely cause in this case the acceleration is constant.

if the acceleration wasn't constant, i wouldn't have used one of the BIG 5 equations nor would i have submitted a reply. Besides, if the acceleration wasn't constant, your method would not have worked as well.

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thermodynamicaldude
yeah...i know. ;-D Most people probably would solve this problem like you did...but I thought it would be interesting to post a method that doesn't involve a big 5 equation derived from calculus...and uses, in a sense, basic mathematics to find acceleration.