# 1-D Kinematics question

1. Jan 19, 2016

### Thewindyfan

1. The problem statement, all variables and given/known data
Anna is driving from Champaign to Indianapolis on I-74. She passes the Prospect Ave. exit at noon and maintains a constant speed of 75 mph for the entire trip. Chuck is driving in the opposite direction. He passes the Brownsburg, IN exit at 12:30pm and maintains a constant speed of 65 mph all the way to Champaign. Assume that the Brownsburg and Prospect exits are 105 miles apart, and that the road is straight.

How far from the Prospect Ave. exit do Anna and Chuck pass each other? x =

2. Relevant equations
Anna's position: 75t + 0
Chuck's position: -65t + 105

3. The attempt at a solution
I came up with the two equations above and then set them equal to each other, getting 140t = 105 => t = .75 hours, which I plugged in to Anna's position equation since it's asking how far away we are from her initial position and got 56.25, but when I enter my answer it reminded me that I forgot to account for the fact that Chuck leaves a half an hour later, and for the life of me I cannot remember how to account for that in my equation for chuck for some reason. That's what I need help with. Thanks!

2. Jan 19, 2016

### Staff: Mentor

Those are expressions, not equations.

Probably the easiest way to visualize this problem (and to answer your question about how to express the time delay) is to make a graph of the positions as a function of time for the two vehicles. Make the horizontal axis of your graph be time, and the vertical axis be the distance from Anna's starting point. Draw the straight tilting lines for the two vehicle positions, and look for where they cross (where the two expressions are equal).

Be sure to start the 2nd vehicle with the time delay that is mentioned. Now do you see how to modify the 2nd expression to account for the delayed start time?

3. Jan 19, 2016

### Thewindyfan

Ah, sorry for the incorrect terminology. It seems not doing much physics work for a week causes me to forget simple things!
Thank you for the tip, I realized I had to think about that after sorting out the information given. I can't believe I forgot such a simple thing!
I ended up revising the expression for Chuck's position to be -65t + 137.5, which I got by solving for the extra distance d since he would be at the 105 mile marker in t = .5 hours. This led to the correct answer, so thank you!