Kinematics: when will the two cars meet?

In summary, two cars start from rest and travel in opposite directions. Car A accelerates at 6m/s^2 to a maximum speed of 30m/s, while Car B accelerates at 5m/s^2 to a maximum speed of 35m/s. There is a distance of 200m between them. They will meet between 5 and 7 seconds, with Car A covering a distance of 105m and Car B covering a distance of 90m after 6 seconds. To find the exact meeting time, the combined speed of the cars must be calculated, taking into account their respective accelerations. Alternatively, the relative motion of the cars can be considered to find the meeting time and distance.
  • #1
get_physical
85
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
How much distance is there to cover yet? What is the combined speed of converging between 5 and 7 seconds?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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.
 
  • #8
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.
 
  • #9
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?
 

1. How is kinematics used to determine when two cars will meet?

Kinematics is a branch of physics that deals with the motion of objects. In the case of two cars meeting, kinematics can be used to analyze the position, velocity, and acceleration of each car to determine when they will intersect or meet.

2. What information is needed to calculate when two cars will meet using kinematics?

To calculate when two cars will meet using kinematics, you will need information such as the initial positions and velocities of the cars, as well as the acceleration of each car. This information can be obtained through measurements or given in a problem.

3. Can kinematics be used to determine the exact location where two cars will meet?

Yes, kinematics can be used to calculate the exact location where two cars will meet. By analyzing the position, velocity, and acceleration of each car, the point of intersection or meeting can be determined.

4. How does the speed of the cars affect when they will meet?

The speed of the cars plays a crucial role in determining when they will meet. If the cars are moving at a higher speed, they will cover more distance in a shorter amount of time, resulting in them meeting sooner. Similarly, if the cars are moving at a slower speed, it will take them longer to meet.

5. Is there a specific formula for calculating when two cars will meet using kinematics?

Yes, there are specific formulas for calculating when two cars will meet using kinematics. These formulas include equations for position, velocity, and acceleration, as well as the equation for solving for time when two objects meet. These formulas can vary depending on the specific scenario and type of motion being analyzed.

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