# A point particle ahead of another

1 vote
47 views

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

edited Jun 25, 2018
The way you had written option D was not clear so I interpreted what I thought is should be and changed it.

1 vote

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 (28,466 points)
selected Jun 25, 2018
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.