# Linear motion of a car problem

1. Jun 5, 2009

### weskerq8

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

A car is stopped at a red traffic light. At the instant the traffic light turns green, the car
starts to move forward with a constant acceleration of 3.20 m/s2. At the same time as the
light turns green, a truck traveling in the same direction with a constant speed of 20.0
m/s, overtakes and passes the car.

How far beyond its starting point does the car overtake the truck?

So, for the car we have:

- initial velocity = 0
- acceleration = 3.20

for the truck we got:

- final velocity = 20.0
- acceleration = 0

2. Relevant equations

x-x. = v. t + 1/2 a t^2
v = v. + a t
v^2 = v.^2 + 2 a (x-x.)

3. The attempt at a solution

I really have no clue on how to deal with this as i don't have time or final velocity for the car, a lot of missing information here and i really don't know how to approach to the answer.

2. Jun 5, 2009

### fatra2

Hi there,

The car has a constant acceleration, therefore no theoretical upper speed limit. To answer questions like these, you need to put the position equation for the two moving objects. One has a constant speed, and the other a constant acceleration.

Then, you can guess the rest, the car will catch the truck when the position of both are the same.

Cheers.

3. Jun 5, 2009

### weskerq8

so according to what you said, my answer going to be 250 meters?

here is my work so you can check if it's right or wrong.

Car: X= v. t + 1/2 a t^2

X = 1.6 t^2

Truck: X = v. t + 1/2 a t^2

X=20.0 t

----

then, x = x

1.6 t^2 = 20.0 t
finally, t = 12.5 s

then i substitute t back to any of the previous equations to get X = 250 m

is that correct?

Thanks btw :)

4. Jun 5, 2009

### fatra2

That's what I find to.

It means that it takes the car 12.5sec to catch the truck. With this, you can also determine the car's speed at this time.

Cheers