Kinematics: when will the two cars meet?

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Homework Help Overview

The problem involves two cars starting from rest and traveling in opposite directions with different accelerations and maximum speeds. The objective is to determine when the two cars will meet, given an initial distance of 200 meters between them.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the distances traveled by each car at specific time intervals and question how to find the exact time of meeting. Some suggest calculating combined speeds and relative motion, while others express confusion about the necessity of combining speeds due to the cars' varying accelerations.

Discussion Status

The discussion is ongoing, with participants attempting to clarify their understanding of relative motion and acceleration. Some have proposed equations based on the distances traveled by each car, while others are questioning the assumptions made about constant speed and acceleration. There is no explicit consensus on the best approach yet.

Contextual Notes

Participants note that after 6 seconds, the cars are still 5 meters apart, which raises questions about how to approach the problem of finding the time until they meet. The varying acceleration of each car and the need to account for the time taken to reach maximum speeds are also under discussion.

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Homework Statement


Two cars start from rest and travel in opposite directions.
Car A travels at 6m/s^2 to a maximum of 30m/s. Car B travels 5m/s^2 to maximum of 35m/s. 200m between them. When will they meet?

Homework Equations


d=vt
d=vit + 1/2 at2


The Attempt at a Solution


at 6 seconds:
car A:
d=vit + 1/2 at2 +30m
=105m

Car B:
d=vit + 1/2 at2 = 90m

5m left in between them. How do I find when they meet?
 
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Clearly the cars will meet between 5 and 7 seconds. Since t = 5 s, car A moves uniformly, car B keeps accelerating. It may be convenient to compute the distance at t = 5 s.
 
At 5s:
Car 1
d= 75m

Car 2
d = 62.5m

Yes, I know they'll be meeting in between 6 and 7 seconds but how do I find the exact time?
 
How much distance is there to cover yet? What is the combined speed of converging between 5 and 7 seconds?
 
get_physical said:
Car A travels at 6m/s^2 to a maximum of 30m/s. Car B travels 5m/s^2 to maximum of 35m/s. 200m between them. When will they meet?
If in t more seconds of travel car A moves a distance "d" in the same time that car B travels a distance "5-d" they will meet at time t at that position d.
 
voko said:
How much distance is there to cover yet? What is the combined speed of converging between 5 and 7 seconds?

62.5m left. what do you mean by combined speed of converging between 5 and 7 seconds?
Why are we looking for the combined speed?
 
NascentOxygen said:
If in t more seconds of travel car A moves a distance "d" in the same time that car B travels a distance "5-d" they will meet at time t at that position d.

No it doesn't work that way because their speed is not constant.
 
get_physical said:
62.5m left. what do you mean by combined speed of converging between 5 and 7 seconds?
Why are we looking for the combined speed?

If you know the combined speed, which is uniformly accelerated, then finding the time to cover 62.5 m is very simple.
 
voko said:
If you know the combined speed, which is uniformly accelerated, then finding the time to cover 62.5 m is very simple.

Ok, I still don't understand why I need to combine the speed. One car is going at 30m/s after 6 seconds and the other is not yet going at 35m/s because it takes 7 seconds for it to reach maximum velocity..
 
  • #10
get_physical said:
No it doesn't work that way because their speed is not constant.
It works even when their speed is not constant.
 
  • #11
get_physical said:
No it doesn't work that way because their speed is not constant.
It's the definition of 'meeting'. They are at the same place at the same time.
 
  • #12
NascentOxygen said:
It works even when their speed is not constant.

But why is it 5-d? car A is accelerating at 6m/s^2 and car B is accelerating at 5m/s^2. At t= 1, they are not at 30m/s nor 35m/s respectively. It takes car A 5 seconds to reach 30m/s and then the speed stays constant. Car B takes 7 sec to reach maximum velocity at 35m/s.

Please help!
 
  • #13
get_physical said:
But why is it 5-d?
Because in the OP you calculated that after 6 seconds they would still be 5m apart. So if the point where they will meet is another d metres from A then it's 5-d from B.
car A is accelerating at 6m/s^2 and car B is accelerating at 5m/s^2. At t= 1, they are not at 30m/s nor 35m/s respectively. It takes car A 5 seconds to reach 30m/s and then the speed stays constant. Car B takes 7 sec to reach maximum velocity at 35m/s.
Why is t=1 interesting? Take it from how things stand after 6 seconds:
5 metres apart
A at constant 30m/s
B currently at 30m/s and still accelerating
You can either do it by relative motion:
What is their relative velocity?
What is their relative acceleration?
How long will it take to close the gap, assuming B's acceleration doesn't change?
​
or by supposing they meet t seconds later, A having traveled d and B having traveled 5-d, and write out the equations for that.
 
  • #14
haruspex said:
Because in the OP you calculated that after 6 seconds they would still be 5m apart. So if the point where they will meet is another d metres from A then it's 5-d from B.

Why is t=1 interesting? Take it from how things stand after 6 seconds:
5 metres apart
A at constant 30m/s
B currently at 30m/s and still accelerating
You can either do it by relative motion:
What is their relative velocity?
What is their relative acceleration?
How long will it take to close the gap, assuming B's acceleration doesn't change?
​
or by supposing they meet t seconds later, A having traveled d and B having traveled 5-d, and write out the equations for that.

oK, I don't really understand relative velocity, so I am going to suppose they meet t seconds later.

For Car B:
D=Vit +1/2 at^2 so D= 5-d, vi= 0 and a= 5m/s^2. I isolated t and equated to Car A.

t=d/30 = sqrt((10-2d)/5) ? Solve for d, then solve for t?
is that the correct way to approach it?
 
  • #15
get_physical said:
For Car B:
D=Vit +1/2 at^2 so D= 5-d, vi= 0 and a= 5m/s^2. I isolated t and equated to Car A.
Think again about vi.
What corresponding equation do you have for car A?
 
  • #16
T= D/30 for car A
 
  • #17
get_physical said:
T= D/30 for car A

OK. What about that vi for car B?
 
  • #18
vi= 30m/s? Ok, I think I got the answer. But it's a bit complicated to do it this way. Can somebody explain the relative velocity way? Thank you
 
Last edited:
  • #19
At times greater than 5 seconds, car 1's speed is 30 m/s, and its distance is 75 +30(t-5) = 30t - 75
At times greater than 5 seconds (but less than 7 seconds), car 2's distance is 2.5 t2. If they meet between 5 and 7 seconds, the sum of these two distances must total 200.
 
  • #20
how do I do it with relative velocity?
 

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