Kinematics: when will the two cars meet?

[get_physical]

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?

 

[voko]

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.

 

[get_physical]

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?

 

[voko]

How much distance is there to cover yet? What is the combined speed of converging between 5 and 7 seconds?

 

[NascentOxygen]

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.

 

[get_physical]

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?

 

[get_physical]

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.

 

[voko]

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.

 

[get_physical]

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..

 

[NascentOxygen]

No it doesn't work that way because their speed is not constant.

It works even when their speed is not constant.

 

[haruspex]

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.

 

[get_physical]

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!

 

[haruspex]

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.

 

[get_physical]

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?

 

[haruspex]

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?

 

[get_physical]

T= D/30 for car A

 

[haruspex]

T= D/30 for car A

OK. What about that vi for car B?

 

[get_physical]

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

 

[Chestermiller]

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.

 

[get_physical]

how do I do it with relative velocity?