# Kinematics qns

1. Jul 13, 2007

### lycheeliang

1. The problem statement, all variables and given/known data
From eqn a=-300v^2
Find time taken for acceleration to change from 1.50ms^-1 to 0.75ms^1

2. Relevant equations
a=dv/dt

3. The attempt at a solution

I tried integrating, but still cannot get the ans.

2. Jul 13, 2007

### JohnDuck

This is a simple differential equation. Try separation of variables.

3. Jul 13, 2007

### Dick

JohnDuck is quite right, except I think the usual term is 'separable' rather than 'separation of variables'. And it is fairly easy. You'll just have to be clearer to us where you are confused.

Last edited: Jul 13, 2007
4. Jul 14, 2007

### HallsofIvy

Staff Emeritus
"separable' is the adjective, "separation" the noun. Yes, this is a "separable" equation and the technique for solving is "separation of variables"- you are both right.

lycheeliang, you have dv/dt= -300v2 so, separating the variables (the verb form!) you have dv/v^2 = -300dt. Integrate both sides of that.

5. Jul 14, 2007

oh thanks!