This topic is 3990 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

ive got my self in a pickle and didnt know where else to turn. The distance, S, meters from a fixed point of a vehicle traveling in a straight line with a constant acceleration a m/s^2, is given by S = ut + ((1/2)at)^2, where u is the initial velocity in m/s and t is the time in secs. determine the initial velocity and the acceleration given that s = 42m when t = 2 secs. and s = 144m when t = 4 secs. find also the distance traveled after 3 secs. im confused big time. ive got two equations 42 = u2 + ((1/2)a2)^2 144 = u4 + ((1/2)a4)^2 but dont know what to do with em. ive been googling for the last 4 hours. thanks for any help.

##### Share on other sites
First off it seems you've got the equation slightly wrong, according to memory (and a book, too) it should read: s = u*t + a*t*t/2 (that's only the t squared, not (a*t/2)^2 ).

You've got information to form two equations which is enough to get your two unknown's, which you did:

s = 42 and t = 2   =>    42 = 2*u + a*2*2/2   =>   42 = 2*u + 2*as = 144 and t = 4  =>   144 = 4*u + a*4*4/2   =>  144 = 4*u + 8*a

There's a couple of ways to solve this but a simple one it to subtract the second equation with the first one times 2 (I choose 2 since that makes the 4*u "disappear" in the second equation):

144 - (42 * 2) = 4*u + 8*a - 2 * (2*u + 2*a)
144 - 84 = 4*u + 8*a - 4*u - 4*a
60 = 4a
a = 15

Now you can put that into the equations to get u:

42 = 2*u + 15*2*2/2
u = 6

Now with a and u getting s after three seconds is trivial.

Edit: I was bothered with what the procedure was called and went googling, found a nice article that explains it: Ask Dr. Math - Substitution and elimination.

##### Share on other sites
thanks very much, nice explenation made it clear for me.

1. 1
2. 2
3. 3
4. 4
Rutin
13
5. 5

• 26
• 10
• 9
• 9
• 11
• ### Forum Statistics

• Total Topics
633694
• Total Posts
3013373
×