# Kinematics of a particle

1. Jan 13, 2004

### 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=4sec. 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) and then substitute back into original equation. But how do you do this.

2. Jan 14, 2004

### himanshu121

If u want to taste calculus here it goes then

Now $$a=\frac{dv}{dt}=\frac{dv*ds}{ds*dt}=\frac{vdv}{ds}=3s^{-\frac{1}{3}}$$

$$vdv=3s^{-\frac{1}{3}} ds$$
integrate u wll get $$v^2=9s^{\frac{2}{3}}+c$$

From conditions given c=0
therefore $$v^2=9s^{\frac{2}{3}}$$
Now $$v=3s^{\frac{1}{3}}$$

v=ds/dt

so we again have

$$s^{-\frac{1}{3}} ds = 3dt$$
Again integrating u get
$$\int s^{-\frac{1}{3}} ds = \int 3dt$$
u get
$$\frac{3}{2} s^{\frac{2}{3}}=3t+c$$

From the given conditions c=0
so we have
$$s^{\frac{2}{3}}=2t$$

So at t=4, s=$$8^{\frac{3}{2}}$$

and hence acceleration a= 3$$8^{-\frac{1}{3}}$$ = 1.06

Last edited: Jan 14, 2004
3. Jan 14, 2004

### jjiimmyy101

Thank-you.

I'll apologize before I even ask.
Sorry.

I know you integrated, but could you show me in more detail how you went from v*dv = 3*s^-1/3*ds
to
v^2 = 9*s^2/3 + c

4. Jan 15, 2004

### himanshu121

It is a basic formula

$$\int x^n dx = \frac{x^{n+1}}{n+1}$$