Originally Posted by **Aladdin** Two cars face each other on a horizontal road.
Car A starts from rest at t=0 and travels with a constant acceleration of 60m/s^2, until it reaches a speed of 80ft/s. Afterward it maintain the same speed.Also after t=50 sec , Car B located 6000 ft down the road is traveling towards A with a constant speed of 60 ft/s.
Determine the distance traveled by A when they pass each other.
We've done this problem , having both cars launched at the same time.
What are the changes that might occur.
Thanks in advance. |

Is the acceleration in m/s^2 or ft/s^2? I'll assume it's ft/s^2.

Let's make a table of information:

Car A: First let's find out where and when car A reaches 80 ft/s. (This will supposedly happen before the two cars pass each other.) Set the origin to be where the car starts and the positive direction be the direction it moves in. t will be when the car gets to that point and x will be the position of the car at that time.

t0 = 0 s......t = ?

x0 = 0 ft.....x = ?

v0 = 0 ft/s..v = 80 ft/s

a = 60 ft/s^2

We need both x and t for the second part. Let's start with x. We don't know t so we need an equation for x that doesn't include time. Thus

v^2 = v0^2 + 2a(x - x0)

This tells use that car A reaches v = 80 ft/s at a position x of 53.3333333 ft. The time that this happens at is (using any equation that we like)

v = v0 + at

gives t = 1.3333333 s.

Car A travels at this speed for 50 s - 1.3333333 s = 48.666666 s. It covers a an additional distance of (80 ft/s)(48.6666666 s) = 3893.333333 ft This puts car A at a position of x0 = 53.33333 ft + 3893.333333 ft = 3946.666666 ft at a time of 50 s.

Now we need to set up the next part...where the cars cross paths. This is going to require information from car A and car B. I will call the position where the cars pass each other X. They will do so at a time T. I will also reset the time to be 0 when car A gets to x = 3946.666666 ft.

Car A:

t0 = 0 s......................t = T

x0 = 3946.6666 ft.........x = X

v0 = 80 ft/s.................v = 80 ft/s

a = 0 ft/s^2

Car B:

t0 = 0 s............t = T

x0 = 6000 ft......x = X

v0 = -60 ft/s.....v = -60 ft/s

a = 0 ft/s^2

Can you take it from here? (Hint: Find two equations in X and T and solve them simultaneously.)

-Dan