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### Two Cars

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Date: 08/11/99 at 07:42:33
From: Sandi
Subject: Kinematics

Hi Doctors,

I am hoping that you can help me with this problem.

Cars A and B are stationary on a straight road, side by side. Car A
moves off with acceleration of 1.20 m/(s^2), maintaining this rate
until it reaches a speed of 108 km/h (30 m/s). It then continues to
cruise at this rate. 15 seconds after A starts, B sets off in pursuit
of A with constant acceleration of 2 m/(s^2) until it draws even with
A.

Find:

a) how long car B takes to draw even with car A (to nearest 10th of a
second), and

b) the distance car B travels, to the nearest meter, before drawing
even with car A.

So far I have worked out that the displacement of the two cars needs
to be the same, so I need two equations that I can say equal each
other to solve for t.

Car A displacement = [(.5)(25)(30)] + 30(t-25)
Car B displacement = [(.5)(15)(30)] + 30(t-45) ... or is it (t-30)?

The part that confuses me is that car B starts 15 seconds later.

I've tried saying that for car B,

V = 2t + 15

then does

s = t^2 + 15t + 450

because when v = 0, t = 15.

Anyway, I know that the answer is 32.3 seconds and I have tried many
different approaches. I tried to make t = 0 when car A reaches 30 m/s
but got even more confused.

Could you please show me how to do this so that any other questions of
this type won't be a problem in the future?

Thank you very much,
Sandi
```

```
Date: 08/11/99 at 15:39:34
From: Doctor Anthony
Subject: Re: Kinematics

>a) how long car B takes to draw even with car A (to nearest 10th of a
>   second)

The time for A to reach 30 m/s is given by

v = u + at,
so

t = (v-u)/a
= (30-0)/1.2
= 25 secs

and distance is given by

s = (1/2)(u+v)t
= (1/2)(30)(25)
=  375 m

Thereafter A will travel 30 m/sec, so if t1 = total time that A has
been moving, we know that the distance is

s_A = 375 + 30(t1 - 25)   ...........................(1)

The total time that B moves until it reaches A is t1-15. The distance
traveled by B is

s_B = 0 + (1/2)(2)(t1-15)^2   .......................(2)

Equating the distances in (1) and (2)

(t1-15)^2 = 375 + 30(t1-25)

(t1-15)^2 - 30(t1-25) - 375 = 0

t1^2 - 30.t1 + 225 - 30.t1 +750 -375 = 0

t1^2 - 60.t1 + 600 = 0

This is a quadratic in t1, which has a positive root 30+10.sqrt(3),
and so

t1 = 30 + 10.sqrt(3)
= 47.32 sec. or 12.68 sec.

and this is the time A has been moving, so the time for B to overtake
A (from when B starts) is

t1 - 15 = 47.32 - 15 sec.   or   12.68 - 15 sec.
= 32.32 sec.        or   -2.32 sec.

We can disregard the latter, and our answer is 32.32 sec.

>b) the distance car B travels, to the nearest meter, before drawing
>   even with car A.

This is the same distance as car A has traveled

s = 375 + 30(t1 - 25)

= 375 + 30[30+10.sqrt(3)-25]

= 375 + 884.847

= 1044.615 meters

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Physics/Chemistry

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