Finding the distance at which Car 1 overtakes Car 2

[azadder]

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
\Deltax=v₀t+1/2at²
v_av=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.

 

[Hootenanny]

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.

 

[azadder]

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]

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 traveled in time t.

 

[PeterO]

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]

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.