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When 2 Cars With Differing Velocities Will Meet

  • Thread starter Kova Nova
  • Start date
Kova Nova
1. Homework Statement
A red car moves with a constant velocity of 20m/s. Exactly 5 minutes later a blue car leaves from the same point with a constant velocity of 30m/s in the same direction as the red car.
a.) For how long (time) does each car travel before meeting?
b.) How far do the cars travel before meeting?
2. Homework Equations
I'm not sure if these are relevant but some equations in mind are:
Vf = Vi + at
x = xi + vit + (1/2a)(t^2)
Vf^2 = Vi^2 +2a(xf-xi)
3. The Attempt at a Solution
5 minutes is equal to 300s for easier conversions. The red car has already reached a distance of 6000m when the blue car finally begins to head out. I have attempted making an XY chart so for every distance I plug in for the red car I get the outcome for the blue car. (e.g. when Red car has traveled 5 minutes, it has gone 6000m but blue car is at 0m; then at 7 minutes Red car has traveled 8400m while blue car has traveled 3600m.) I did this all the way to 13 and 14 minutes where the blue passed by the red at 14 minutes so the answer has got to be between 13 and 14 minutes. However, I am looking for a more precise method of figuring out the answer, such as an equation to find the exact time if there is one. Much thanks to anyone who can help!
 

Answers and Replies

MarkFL
I would let ##d_R## be the distance traveled by the red car and ##d_B## be the distance traveled by the blue car. If the red car travels for ##t## seconds, how long does the blue car travel? Using the relation ##d=vt##, can you give an equation for each car? What relation do the two distances share?
 
CWatters
Science Advisor
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When you do it graphically the point where the two lines cross represents the solution to the simultaneous equations. So as Mark say, write your simultaneous equations and solve them. You might think there are too many variables but remember that for two cars to pass each other they must be at the same place at the same time.
 
shiv222
Suppose cars meet time t after blue car starts. Distance traveled by both cars after time t is same.
Distance traveled by red car is = (300+t)X20
That by blue car is = 30t
As both distances are equal:
(300+t)X20 = 30t
We get t = 600 sec is the time for two cars to meet.
 

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