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A point particle ahead of another

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Two vehicles move one after the other with velocities $v_1$ (the one ahead) and $v_2$ (the one behind). The driver of the second vehicle( at a distance d from the first one) slows down so as to avoid reaching it, thus it suffers a deceleration of modulus a. Which ONE of the following statements is correct?
A) If $v_2-v_1= \sqrt{2ad}$ the second car will reach the first one.
B) Only if $v_2-v_1 > \sqrt{2ad}$ will the second car reach the first one.
C) Only if $v_2-v_1 < \sqrt{2ad}$ will the second car reach the first one.
D) If $v_2-v_1 \gt 0$ there is no value of deceleration $a$ which avoids the meeting between vehicles.

This is what I tried:

As the second car slows down so as not to reach the leading car: $v_{2}>v_{1}$

The velocity of the car going behind is (it is not constant as it is decelerating):

$V_{B}= V_{2} - at$

So as you can see I stated that car2 catches car1 when $V_{B} = V_{1}$ but to be honest I did it because I knew I was going to obtain b) if I did so. So if it is like that, could you explain me why?

Thanks

asked Jun 25, 2018 in Physics Problems by Jorge Daniel (606 points)
edited Jun 25, 2018 by Jorge Daniel
The way you had written option D was not clear so I interpreted what I thought is should be and changed it.

1 Answer

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Best answer

You are correct that car #2 must have a speed less than $v_1$ when it reaches car #1. This is because the relative speed is then zero. If car #2 had a speed greater than $v_1$ at this instant, it will still be moving faster than car #1 at the next instant, so the cars will collide.

It is easier to see what is happening by using a frame of reference in which one car (eg
#1) is stationary. Then the following car (#2) starts with velocity $v_2-v_1$ and is initially distance $d$ from car #1.

Car #2 must decelerate and reach a velocity of zero within a stopping distance $s$ which is equal to or greater than $d$ in order to reach car #1. Using the kinematic formula $v^2=u^2+2as$ we obtain $$0=(v_2-v_1)^2-2as$$ $$v_2-v_1 = \sqrt{2as} \ge \sqrt{2ad}$$ because $s \ge d$.

Option A is correct because B does not include the possibility $v_2-v_1=\sqrt{2gd}$ which also allows car #2 to reach car #1.

answered Jun 25, 2018 by sammy gerbil (28,466 points)
selected Jun 25, 2018 by Jorge Daniel
Ok Sr let me see if I got why $v_2-v_1$. I will explain myself using vectors. Firstly we draw an arrow pointing to the right direction which is not going to change in function of time, which we name $v_1$ (stationary). Secondly we draw another arrow pointing to the right as well. This is larger in length than $v_1$ at t=0 and it gets shorter and shorter in function of time (constantly decelerating). Thus there is an instant in time where $\frac{dx}{dt}= v_2-v_1$ right? Sorry if I am justifying things that may appear so trivial but I may be asked in future tests for discussing why, and I want to be sure.
After poring over your solution, in my opinion the correct answer is a) as b) states only if and it also reaches the car when $v_2-v_1= \sqrt{2ad}$
Yes if $x$ is the distance between the cars then $\frac{dx}{dt}=v_B-v_1$. ($v_B$ is variable, $v_2=v_B(0)$ is constant.)  The distance between the cars stops changing when $\frac{dx}{dt}=0$ ie when $v_B=v_1$.

Your reason for rejecting option B might be correct but it is pedantic. There is some doubt about what the question is asking for : the question mentions that car #2 wants to "avoid overtaking" car #1, whereas the answers ask about car #2 "reaching" car #1. Are these intended to mean the same thing? If $s=d$ then car #2 reaches but does not overtake car #1. If $s\gt d$ then car #2 reaches and overtakes car #1.

Physics questions are not good if the difference between options depends on fine distinctions in meaning ("reaching" vs "overtaking"), rather than clear differences in physical effects. There is a fine line between A and B, which makes the question a poor one in my opinion.
My mistake, I should not have used the word overtake. The original question asks for reaching only. Changed now. Then, the answer is b) right?
No, I think this makes it clearer that the intended answer is A, for the reason you gave. I will change my answer.
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