# Kinematic word problem

1. Jan 24, 2007

### physnotmything

Hi everyone, i have a exam coming up for Physics and this question from the kinematic unit is really bothering me. I've tried many times but i have no idea how to solve it. Some help would be greatly appreciated, thanks.

1. The problem statement, all variables and given/known data

Question: Town A and Town B are 6.0 x 10^2 (600) meters apart. Kevin leaves Town A and heads for Town B at a constant speed 30m/s. Peter leaves Town B at the same time and heads for Town A at an acceleration 1.5 m/s^2 from rest.

a) at what distance from Town A will they meet each other?
b) How much time passes before they meet?

2. Relevant equations
I used only this equation so far
Delta d = (Velocity 1)( delta time) + 1/2 (acceleration) (delta time)^2

3. The attempt at a solution

I've stated some of the unknowns from Kevin and Peter, and found only the following.
Kevin-
Velocity 1 = 30 m/s
displacement Kevin = Displacement Peter

Peter-
acceleration = -1.5
velocity 1 = 0
displacement Peter = Displacement Kevin

Using those information i tried connecting the two equations with the same unknown variable. But I only got one equation, and it was for Peter.

delta D = v1(t) + 1/2 (a)(t)^2
= 0(t) + 1/2(-1.5)(t)^2
= -0.75(t)^2

2. Jan 24, 2007

### hage567

I don't get why your acceleration for Peter is negative, if he is accelerating towards Town A. Also, I am confused by your displacement Kevin = displacement Peter. You can put the distance Peter travels in terms of the distance Kevin travels, since you know the separation between A and B. I'm not sure which unknown you were trying to connect them with, but find an expression for the distance each person travels, and look at what variable would be the common one. Base your substitution on that, and solve the resulting equation. (I got a quadratic )

Hope that helps.

3. Jan 25, 2007

### Chi Meson

The quadratic seems to be unavoidable.

IT makes sense to me to solve for t first.