# Linear Kinematics problem

1. Oct 14, 2009

### andysmax

1. The problem statement, all variables and given/known data

Two cars are at the start lines of two roads at right angles to each other and pointing to the intersection 500 metres away. Car A starts from rest with uniform accelereation of 0.5m/s2 and accelerates for 20 seconds. It then continues at the speed reached. Car B starts 5 seconds later and 20 metres further away with an acceleration of 0.7m/s2.
Calculate wrt the start time of Car A:

a) Time taken for B and A to be the same distance from the cross-road.

b) The distance from the cross roads at the above time

c) The average velocity of Car A at this distance.

2. Relevant equations

v=u+at
v2=u2+2as

3. The attempt at a solution

First i used v=u+at and plugged the results into a table showing velocity against time for car A and B, for upto 60 seconds at ten second intervals.
Then i transposed the second equation for s to work out the distance s=v2-u2$$/2a$$ for car a and s=(v2-u2$$/2a$$)-20 for car B.
This was then plugged into a second table showing distance against time.
From this table it was deduced that the area of interest was between 50 and 60 seconds.
I carried on with this process until i narrowed it down.
The answer i got for a) was 57.67 seconds.
b) was just working the time back through the previous equations which gave 23.3m (500-476.7).
c) i calculated change in distance over change in time: 476.7/57.67 = 8.267m/s2

However, this was time consuming and i am wondering if there was a simpler way to tackle this(if indeed the above solution is correct!), possibly as a simultaneous equation?

Any help is greatly appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 14, 2009

### ideasrule

Rest assured that you don't need to use tables for any physics problem. When the two cars are at the same distance from the cross-road, two things are the same for both cars: distance travelled and time taken. Which equation relates distance, acceleration, and time? Start there.

3. Oct 14, 2009

### Delphi51

Your method looks pretty good. I did it a different way and got a different answer! Kind of complicated and easy to make a little mistake, which I have a habit of doing.
For car A I did the first 20 seconds as v = at = .5*20 = 10
d = .5*a*t^2 = 100 m
Then for times larger than 20 we have d = 100 + vt = 100+10(t-20)

For car B, d = do + vi*t + .5*a*t^2
d = -20 +.5*.7*(t-5)^2

for (a) I set the two distance equations equal, collected like terms and got this quadratic: .35*t^2 - 13.5*t + 88.75 = 0
It has two solutions, one less than 20 so no good, and t = 30.17 s
I checked the distances at that time and they are equal (if my equations are correct!).

4. Oct 14, 2009

### andysmax

Car B starts 20 m further away. That's why i went the table route.

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