# Motion problem. Acceleration with respect to displacement

• Feb 9th 2008, 04:18 AM
chrsr345
Motion problem. Acceleration with respect to displacement
Im having a problem solving a physics related problem. This is it:
A truck travels at a speed of 4m/s along a circular road with radius if 50m. For a short distance, from s=0, its speed is then increased by a(tan) = (0.05s) m/s, where s is in meters. Determine the speed at s=10.

I thought this was a simple integration to find velocity, but i realized that i cant just integrate with respect to s.There is no time in this problem. So from there, I have no idea what to do really. It is hard for me to visualize this sort of equation. And just to note, a is tangential acceleration. The answer is 4.58 m/s. I just cant get that answer.
Any help is appreciated, thanks!

Chris
• Feb 9th 2008, 09:43 AM
CaptainBlack
Quote:

Originally Posted by chrsr345
Im having a problem solving a physics related problem. This is it:
A truck travels at a speed of 4m/s along a circular road with radius if 50m. For a short distance, from s=0, its speed is then increased by a = (0.05s) m/s, where s is in meters. Determine the speed at s=10.

Please repost this using the wording in the actual problem.

RonL
• Feb 9th 2008, 10:55 AM
chrsr345
That is the wording in the actual problem. it also asks for the magnitude of acceleration after it asks for velocity at s=10. But i can easily find that once i find the velocity. But yes, those are the words.
Like i stated, the accel given is the tangential a, or a-sub-t.

Ive asked a friend who has taught upto diff equations and even he didnt know how to do it (Lipssealed)
• Feb 9th 2008, 04:22 PM
chrsr345
The only forumla i have that even relates to this is a(tan)ds = v dv
So i integrated both sides and got 0.025s^2. at s = 10, i get 2.5m/s + 4m/s = 6.5m/s: Wrong!

Im at a dead stop here.
• Feb 10th 2008, 05:30 PM
ra2warroom