A physics problem related to acceleration

I know this seems like a relatively simple problem, but I'm in the eleventh grade.
I actually did solve this sum but when we checked the answers in class I got it incorrect. I'm quite sure I got it correct anyway here it is:-

The graph

Car A travelling at a constant speed of 40 m\s overtakes car B at t=0. In order to catch up with car A, car B accelerates uniformly for 20sec to reach a constant speed of 50m\s

Calculate
a)How far A travels during the first 20sec:- 800m
b)The acceleration and distance of travel of B during first 20sec:- 1.25m\s2, 750m
c)The additional time taken by B to catch up with A:- I say the answer is 5sec the teacher says it is 0. Which is it??

EDIT: Bah, graph got messed up, please try anyway.

Have you asked the teacher?

If part A and B were correct then there is no way part C can be zero (car A travels 800m, B travels 750m).

I would find it quite odd that the 3rd quesion would even be asked if after solving for part B you would have found that both cars traveled identical distances, ie. 800m (assuming C should in fact be zero).

Sounds like an innocent mistake to me.

My teacher seems incompetent. Is the answer 5 anyway? I need an answer if I'm gonna have to bring it up...

Hi there, I may be reading it wrong but it seems that after accelerating for 20s car B has caught car A - hence they are at the same point.
The equations for constant speed and acceleration can easily be derived from:
distance = speed * time and as I recall look like this:

v = u + at, s = 1/2*(u+v)t, s = ut + 1/2*at^2.

The more I look at this the more suspicious it looks - please reprint full question and clarify if after 20secs car B has caught car A. We know it has higher velocity by then so what else can catch up mean?

Moseley, you seem to be my Physics teacher. The question says in order to catch up with A, B accelerates for 20s to reach a speed of 50m\s. Nowhere does it say that it reaches A after the stipulated time. Also as an added note, I have found the distance moved by the cars after 20s. A is 800m and B is 750m. How can B catch up with A unless its moved the same distance as A.

I am not a physics teacher, but have known a few. Language is always where complications arise.
What, to you, is the difference between 'catches up' and 'reaches'? As far as I am concerned these phrases mean the same thing where position is considered. If velocity is considered then 'catches up' means attains same velocity, however we know that in this instance it has attained a higher velocity, and has not only caught it up but overtaken it.
So to make sense the two cars must be at the same position after 20s, although the acceleration of car B (or velocity of A) would need to be different from those quoted to achieve this.
These are the reasons why I am suspicious of the wording of this question and the answers given.

What do I do to make you understand?



IN ORDER to catch up, not to catch up, IN ORDER. It doesn't mean that it has. It just means that Car B wants to catch up with A and in order to do so it accelerates for 20s.



Read the question please. IN ORDER to catch up, not TILL IT CATCHES UP

Hi FP,
If i understand your question right, 5s is correct. The problem comes down to a very simple delta V=10m/s, delta D=50m, so delta T=5s. (V=D/T).
Hope this helps,
Ron

The way I understand part C. They want you to calculate the time it takes for B to be equal to A after the 20 seconds have passed.

Car A, 40 m/s travels from x=0 to x=800 in 20s
Car B, assuming v0=0 and using 20s to reach 50 m/s (a= v/t =2.5 m/s2) traveling for 20 seconds from x=0 to x=x0 + v0*t + 0.5 a t^2 = 0.5 * 2.5 * 400 = 500 m

that would result in 300 + 40*t = 50*t or t=30 s

So, I wonder where I went wrong in this, Been out of this kind of calculations for over 10 years now.

To get the answer 0, Car B should have had an initial speed when Car A passed it.

Ahh, the Graph: 25 m/s makes a = dV/dt = 25/20 = 1.25 m/s2

x=0+25*20+0.5*1.25*400=500+250=750 m

That would result in 50 + 40*t = 50*t or t=5 seconds.

So, the initial speed as indicated in the graph at first sight does not give the answer 0.

At 30 m/s: a= dV/dt = 20/20 = 1 m/s2
x=0+30*20+0.5*1*400=600 + 200 = 800 m

In which case the teacher would have been right and t=0, (40t=50t, 0=10t, t=0)

So, I hope this helps some.


As a side note, can I colect the speeding ticket of car B, 180 km/h is kinda past most commercial car speed limits.