When 2 Cars With Differing Velocities Will Meet

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AI Thread Summary
The red car travels at a constant velocity of 20 m/s and has a 5-minute head start, covering 6000 meters before the blue car, which travels at 30 m/s, begins its journey. To find when the two cars meet, the distances traveled by both cars must be equal. Setting up the equation (300+t) * 20 = 30t leads to solving for t, which results in 600 seconds for the blue car's travel time. Therefore, the total time for the red car before they meet is 900 seconds, and the distance each car travels can be calculated from these times. This approach provides a precise method for determining the meeting point of the two cars.
Kova Nova

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?

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)

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!
 
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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?
 
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.
 
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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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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