# Finding the distance at which Car 1 overtakes Car 2

Problem 1:
Two cars are traveling along a straight road. Car A maintains a constant speed of 86 km/h; car B maintains a constant speed of 106 km/h. At t = 0, car B is 30 km behind car A. How far will car A travel from t = 0 before it is overtaken by car B?

Problem 2:
At t = 0, a stone is dropped from a cliff above a lake; 1.5 seconds later another stone is thrown downward from the same point with an initial speed of 31 m/s. Both stones hit the water at the same instant. Find the height of the cliff.

These are two problems that came up on the homework, and I still am unable to solve them. I am able to get all the problems with one object involved, I just don't seem to comprehend how to set these up and solve them. Any help is appreciated.

Equations in this chapter:
v = v0+at
$\Delta$x=v0t+1/2at2
vav=1/2(v0+v)

I have tried graphing the problems to understand them, but I think I am lacking the fundamental understanding of what each should look like. I have reread the chapter and I am still in the same boat. Text book: Physics for Scientists and Engineers 5th ed.

## Answers and Replies

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
Welcome to Physics Forums.

HINT: When the two cars/stones overtake/hit the water, their relative displacements will be zero. In order words you are looking for the distance when their two positions are the same.

Thanks for the hint and the welcome!

For the first problem, I am assuming the acceleration is 0. Therefore, I can use the 2nd formula as delta x = (v0)t

I believe delta x of car A would be 86km/hr (t) and delta x of car B would be 106 km/hr (t). To find where they have the same displacement, I am guessing I set it up as 86km/hr (t) = 106km/h (t) + 30km
Solve for t, t = 30 km / (86km/hr - 106 km/hr).

Seem to be the right track?

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
Thanks for the hint and the welcome!

For the first problem, I am assuming the acceleration is 0. Therefore, I can use the 2nd formula as delta x = (v0)t

I believe delta x of car A would be 86km/hr (t) and delta x of car B would be 106 km/hr (t). To find where they have the same displacement, I am guessing I set it up as 86km/hr (t) = 106km/h (t) + 30km
Solve for t, t = 30 km / (86km/hr - 106 km/hr).

Seem to be the right track?
Yes. Spot on. Once you have the time, you then only need to work out how far car A has travelled in time t.

PeterO
Homework Helper
Thanks for the hint and the welcome!

For the first problem, I am assuming the acceleration is 0. Therefore, I can use the 2nd formula as delta x = (v0)t

I believe delta x of car A would be 86km/hr (t) and delta x of car B would be 106 km/hr (t). To find where they have the same displacement, I am guessing I set it up as 86km/hr (t) = 106km/h (t) + 30km
Solve for t, t = 30 km / (86km/hr - 106 km/hr).

Seem to be the right track?

Not quite: car A is the one with the 30km head start. Your "equation" has the advantage with Car B

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
Not quite: car A is the one with the 30km head start. Your "equation" has the advantage with Car B
Good catch! I didn't spot the sign error.