Question

Two cars are racing. They both start at the same point. However, Car B leaves the starting line 1.00 s after Car A. Car A moves with a constant acceleration of 3.50 m/\(s^2\), while Kathy maintains an acceleration of 4.90 m/\(s^2\). Find (a) the time at which Car B overtakes Car A

Answer 1

Suppose B overtakes A after 't' seconds after A starts
Distance travelled by A=(3.5*t^2)/2
Distance travelled by B=(4.9(t-1)^2)/2
Both should be equal. Find t by equating both.

Answer 2

how did you get that again?

Answer 3

@mani_jha could you just explain the logic of that for me? :) did that come from a formula?

Answer 4

B started after A, but with a greater acceleration than that of A. So, B will be behind A for some time, but after a time 't' B will catch up with A. To catch up with A at time t, the distances travelled by both should be equal. That's what I've done there. The distance travelled by a body with acceleration a and time t can be calculated by:
\[s=ut+at ^{2}/2\]
Here u=0, because the players start from rest.
Ok now?

Answer 5

u is what? speed? and s is distance right?

Answer 6

yes, u is the initial speed(which is zero) and s is distance.

Answer 7

hmm oh yeah! thanks so much! i'll try solving..

Answer 8

the time for car B do i write it as t-1? because it started 1 second after Car A

Answer 9

i think this is what i originally did and it gave me an answer different from the one on my book o.O

Answer 10

Yes, the time for B should be (t-1)

Answer 11

im gonna use quadratic formula right?

Answer 12

yes. what answer do you get?

Answer 13

i got 9.42 seconds..my book says something like 5 seconds

Answer 14

I did the calculation and got 6.45 s. You sure you did no mistake?

Answer 15

\[4.9t^2 - 9.8t + 4.9 = 3.50t^2\]
\[1.4t^2 + 9.8t + 4.9 = 0\]
\[14t^2 - 98t + 49 = 0\]
then i took away 7

\[2t^2 - 14t + 7 = 0\]

then quadratic

right?

Answer 16

Yes, right. That gives t=6.45

Answer 17

\[t = \frac{-(-14) \pm \sqrt{14^2 - 4(2)(7)}}{2(2)}\]

right?

Answer 18

exactly

Answer 19

hmm i got 6.46 now... but still not 5 something o.O

Answer 20

Yeah, 6.45 or 6.46 seems to be right, (depending on where you rounded off for the approximation).
Actually, t=0.55 seconds should theoretically be another solution to this, by solving the quadratic.
Now, wherein does the trouble lie, Mr. @lgbasallote ?

Answer 21

the book says 5.46 s

Answer 22

ohkay. you book. hmm.

since car B started 1 second after car A, and the 6.45 seconds that we are getting is the absolute time, starting with car A's movement, so relative to car B, the time after which the over take takes place would be 5.46 seconds.

That's^ how I'd explain the book's answer.

Answer 23

wait...5.46 s is the perspective of car A or B?

Answer 24

car B ofcourse.

A second less, since it started a second later.

Answer 25

so 5.46 seconds after car B goes off...it overtakes car A?

Answer 26

yeah..!

Answer 27

ooohh great...so why did the answer come out with respect to car A? any reason for that?

Answer 28

The answer to that lies with the author of your book.

What we are doing is basically just 'justifying' the answer printed in your book, it may as well be printing error on their behalf.

Answer 29

:P

Answer 30

no...it makes sense...im asking why when solving it what comes out as time is Car A's time and not Car B

Answer 31

i think it lies with the math

Answer 32

Well, first of all the question should state clearly what they want, if not stated otherwise, we assume absolute time since the situation kicks into action.

Answer 33

oh never mind i got it...we looked for t...which was car A....car B was t - 1 im so slow lol

Answer 34

You speedy!