Kinematics of a Particle

[jjiimmyy101]

Problem: A car starts from rest and moves along a straight line with an acceleration of a=(3s^-1/3)m/sec^2, where s is in metres. Determine the car's acceleration when t=4s. ANS: 1.06 m/sec^2

Alright...I know nothing about integrals...really, nothing. I was never taught anything about integrals even though I've taken calculus courses before.

Here's what I think I should do.

Take the equation a = d^2s/dt^2 and INTEGRATE it to find the position(s). But how do you do this.

[NateTG]

Originally posted by jjiimmyy101
Problem: A car starts from rest and moves along a straight line with an acceleration of a=(3s^-1/3)m/sec^2, where s is in metres. Determine the car's acceleration when t=4s. ANS: 1.06 m/sec^2

Here's what I think I should do.

Take the equation a = d^2s/dt^2 and INTEGRATE it to find the position(s). But how do you do this.

If you were trying to find the position, then that would be correct. Since you're trying fo find the acceleration, you can just use the formula.

[jjiimmyy101]

what formula?

[NateTG]

My bad, I thought s was in seconds.

I don't see an easy way to deal with that one if you don't know how to integrate.

[jjiimmyy101]

How do you integrate?

[eddo]

Are you from the U of S? Because i was working on that exact problem before I came on here, very strange. Anyways, as was said it can't really be done without integrating, which if you are from the U of S, they haven't taught us yet. But we have learnt antiderivatives, which should help you. Start with the relationship ads=vdv (which you can get by eliminating the dt term in a=dv/dt and v=ds/dt). Solve for a to get a=vdv/ds and substitute this into the equation given in the question. Now get the v and dv on the same side, as well as the s and ds terms. It should look something like this:
vdv=3s^(-1/3)ds. Integrate both sides, which basically means to take the antiderivatives. This leaves v and s: v^2=9s^(2/3). Solve for v, and than substitute ds/dt for v. Once again get the s and ds on the same side, and dt on the other side, and integrate again (antiderivative). You now have s as a function of t! From here you can either substitute in t=4 and find s, than put this s value into the original equation to get a, or you could find the second time derivative of s to get an expression for a as a function of t, than put in 4 for t. Both will give you the same answer. Hope this helped, although I'm sure it's confusing to follow.

[himanshu121]

I believe U have also posted the same Question in maths section So look at my reply which is similar to that of eddo