SOLUTION: I don't know if this is the right topic? Question: A stationary car, a, is passed by car b moving with a uniform velocity of 25m/s. Five seconds later, car a starts moving with

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: I don't know if this is the right topic? Question: A stationary car, a, is passed by car b moving with a uniform velocity of 25m/s. Five seconds later, car a starts moving with       Log On


   



Question 997244: I don't know if this is the right topic?
Question: A stationary car, a, is passed by car b moving with a uniform velocity of 25m/s. Five seconds later, car a starts moving with a constant acceleration of 2m/s^2 in the same direction.
Construct a pair of simultaneous equations and plot on a graph using a spreadsheet to find how long it will take car a to draw level.
I have tried to do this question but I don't know how to work out a pair of simultaneous equations I have got a quadratic t%5E2%2B15t%2B25. Not sure how to put that into a spreadsheet to make a graph. I used the equation s=ut%2B%281%2F2%29at%5E2 and substituted the values from car a and car b in and came up with 2 formulas s=%281%2F2%292t%5E2 for car a and s=25t I then made them equal to each other and subtracted 5 seconds of car a making it s=%281%2F2%292%28t-5%29%5E2 due to it starting 5 seconds later. This is how I came to the equation t%5E2%2B15t%2B25

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You almost solved it, but not the way the problem specified.
With t= time in seconds from the moment car B passes car A
The distance from the point where car B passed car A is
s%5BB%5D=25t for car B (for t%3E=0 , because we do not know what car B was doing before), and
s%5BA%5D=%281%2F2%29%2A2%2A%28t-5%29%5E2=%28t-5%29%5E2=t%5E2-10t%2B25 for car A for t%3E=5 .
(Of course, for 0%3C=t%2C5 s%5BA%5D=0 because car A was not moving then).
So when A is level with B again, s%5BA%5D=s%5BB%5D , and
t%5E2-10t%2B25=25t--->t%5E2-10t%2B25-25t=0--->t%5E2-35t%2B25=0 .
The solutions to that equation, using the infamous quadratic formula are
.
Since %2835+%2B-+sqrt%281125%29%29%2F2%3C5 , it is not a solution to our problem.
The solution to our problem is
t=%2835+%2B+sqrt%281125%29%29%2F2%3C5=+about34.27
(You had a little problem with %22%2B%22 and %22-%22 signs, but I make those mistakes sometimes, so that proves that we are both human).

WHAT THE TEACHER WANTS (I think):
Here is how I would answer if it was my homework.
t= time in seconds from the moment car B passes car A.
s%5BA%5D=%281%2F2%29%2A2%2A%28t-5%29%5E2=%28t-5%29%5E2=t%5E2-10t%2B25 for t%3E=5 is the distance traveled by car A since B passes it.
s%5BB%5D=25t (for t%3E=0 is the distance traveled by car B after it passes car A.
So, for t%3E=5 , the system of simultaneous equations that describes the situation in the problem is
system%28s=t%5E2-10t%2B25%2Cs=25t%29 .
I used Microsoft Excel to tabulate and plot as shown below: