Solving for the minimum speed a car must have to catch up to another car?

In summary, the green car can maintain a minimum constant speed of 7.2 m/s to catch up to the blue car, which is accelerating at 0.9 m/s2. The minimum speed is found by setting the derivative of the velocity function to 0 and solving for the time when the cars are at the same position. This occurs at t = 8 s, so the green car must maintain a speed of 7.2 m/s for at least 8 seconds to catch up to the blue car.
  • #1
PastaTapestry
5
0
1. A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.9 m/s2. A green car arrives at the position of the stop-light 4 s after the light had turned green. What is the slowest constant speed which the green car can maintain and still catch up to the blue car?
Answer in units of m/s

So, we're given that ##a = 0.9 m/(s^2)##, ##v_0 = 0##, and that displacement at t = 0 is 0 (all of this for the blue car, or ##C_B##)
For the green car(or ##C_G##), we're given that the car is holding a constant speed, so there must be no acceleration. If there's no acceleration, then the final and initial velocities must be equal for ##C_G##. Additionally, we're given that its displacement = 0 at t = 4 (since it started 4 seconds late).



2. We know that the velocity for ##C_B## is $$v=0.9t$$ We know that the displacement is equal to $$0.45t^2$$
For ##C_G##, we know that ##v## is a constant c, that acceleration = 0, and that the displacement is represented by the function ##vt - 4v## (I believe).

3. I figured that we were looking for the minimum speed that ##C_G## would have to possesses to cause the displacement of both ##C_G## and ##C_B## to be equal, because this would represent that ##C_G## had caught up to ##C_B##. However, I kept getting it down to things like $$0.45t^2 = vt - 4v$$ which you can't solve to my knowledge. So I tried a graphical method. I know that the displacement of ##C_B## is a parabola, and that the displacement of ##C_G## is a linear equation that passes through the point (4,0) [if we graph displacement as y and time as x). So we're looking for a tangent that passes through the point (4,0). However, we don't know the time or displacement this occurs at.

Sorry if this is all too verbose and such, first time posting here so I apologize!
 
Last edited:
Physics news on Phys.org
  • #2
You need the other simultaneous equation. :D

You forgot that this is a minimization problem ... you have an equation with two unknowns: that means there are are a range of possible values that v can take for different times to catch up. You want to find the minimum value out of this range ;)
 
Last edited:
  • #3
Hi Pasta - as a fellow new guy - you're going to LOVE PF!
 
  • #4
PastaTapestry said:
$$0.45t^2 = vt - 4v$$


For real root,
b2>4ac

v2>4(0.45)4v
 
  • #5
Thank you Simon! When you refer to the other simultaneous equation, what are you referring to? I could only find one equation that relates them. I was thinking of using ##v_B = 0.9t##, but I didn't think that would work because ##v_G## and ##v_B## are different, so I'd still be left with two variables.

And aziz, what do you mean? I see what you're doing, but I don't follow where to go from there. In its most basic form it seems to be saying that ##(v_G)^2 > v_G##, which only tells me that ##v_G > 1##. Or am I misreading your post? Sorry!

And thank you enosis! I hope so. :]
 
  • #6
azizlwl said:
For real root,
b2>4ac

v2>4(0.45)4v

v>7.2m/s
So the minimum velocity is 7.2m/s.
Less than 7.2m/s the green car couldn't catch the blue car.

There is no problem if the green car's velocity higher than 7.2m/s
 
  • #7
azizwizi is giving you the method from understanding quadratics ... I'll leave that to him.
He's basically saying that you have a relation between v and t in which not all values of v are possible ... so you can investigate that relation to find out what the smallest possible value of v is :)
PastaTapestry said:
Thank you Simon! When you refer to the other simultaneous equation, what are you referring to?
You have not used one of your constraints - namely, that the speed has to be a minimum.

You found v(t) ... that's equation 1.
The variable t in this equation was the amount of time it takes to catch up.

v(t) is a minimum for catch-up time t when dv/dt=0. ... that's equation 2.

for smaller catch-up times, v is bigger, for longer catchup times v is smaller
... though, there will be some v with no corresponding t because you are going too slow to ever catch up.

I could only find one equation that relates them. I was thinking of using ##v_B = 0.9t##, but I didn't think that would work because ##v_G## and ##v_B## are different, so I'd still be left with two variables.

Imagine: if you are in the green car, and you are going fast, then you will catch up to the blue car at time ##t_1## ... then overtake it, but the blue car is always accelerating ... so it will start gaining on you and overtake at time ##t_2##. Therefore there are two times in which you and he are level with each other. Each time, you and he draw level, you will be doing different speeds ##|v_b(t_2)-v_g|=|v_b(t_1)-v_g| > 0##.

If you go faster ... then those times will get further apart. The difference in your speeds when you draw level will also be greater.

If you go slower, however, the two times get closer together. The difference in your speeds also gets smaller. It should be possible to find a speed where the two times are the same ##t_2=t_1=t## and, therefore, the difference in speeds when you catch up is zero! ##v_g(t)=v_b=v##

If you go slower than this - the blue car will never catch the green car.

So now you know what time the cars meet up for the slowest pursuit-speed that is capable of this you can do the rest.

If you draw the v-t diagrams, on the same axis, for this situation - you will quickly see that the time they meet has to be t=2x4=8s: do it and see!
 
  • #8
I see how both you guys came to your methods/answers now. Thank you! Made this so much easier. :]
 
  • #9
No worries.
These more open questions are closer to what you end up doing in real life ... the fun part is that there are usually many ways of thinking about them.
The take-away lesson here would be that you don't always get just one answer ... you have to find some way to select between the ones that you have.
 

1. What is the minimum speed a car must have to catch up to another car?

The minimum speed a car must have to catch up to another car depends on the initial distance between the two cars, the acceleration of the chasing car, and the velocity of the car being chased. It can be calculated using the formula v = (d + at)/t, where v is the minimum speed, d is the initial distance, a is the acceleration, and t is the time taken to catch up.

2. How does the acceleration of the chasing car affect the minimum speed?

The acceleration of the chasing car directly affects the minimum speed it needs to catch up to the other car. The higher the acceleration, the lower the minimum speed required. This is because a higher acceleration allows the car to cover the initial distance between the two cars in a shorter amount of time.

3. What happens if the initial distance is too large?

If the initial distance between the two cars is too large, the minimum speed required to catch up may be too high for the chasing car to achieve. In this case, it may be impossible for the car to catch up to the other car unless the initial distance is decreased or the acceleration is increased.

4. Does the velocity of the car being chased affect the minimum speed?

Yes, the velocity of the car being chased does affect the minimum speed required for the chasing car to catch up. If the car being chased is traveling at a higher velocity, the minimum speed required for the chasing car will also be higher. This is because the chasing car needs to match the speed of the other car before it can start to catch up.

5. Can the minimum speed be calculated if the time taken to catch up is known?

Yes, the minimum speed can be calculated if the time taken to catch up is known. The formula for this is v = d/t, where v is the minimum speed, d is the initial distance, and t is the time taken to catch up. It is important to note that this formula only works if the acceleration is constant.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
625
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
7K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
809
Replies
1
Views
931
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top