Thread: 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

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

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

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.

use ads = vdv

int=integral

int(.05s)ds = int(v)dv

(.05s^2)/2 = (v(final)^2 - v(initial)^2)/2

plug in 10 and get v(final) = 4.58m/s