SOLUTION: Pls. help me with this. There are so many given I don't know what to do with all of it or what to use. A drag racer accelerates uniformly from rest straight track at 12m/s^2 for

Question 148805: Pls. help me with this. There are so many given I don't know what to do with all of it or what to use.
A drag racer accelerates uniformly from rest straight track at 12m/s^2 for 5.5s, maintains the speed it has reached for 3.0s, and slows down to a stop at the rate of -14m/s^2. What is the total distance traveled by the racer?
Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
The easiest way to approach this is to look at it as three steps of a bigger problem.
The general formula that relates distance to acceleration and speed is
where a is acceleration, v[0] is initial velocity and d[0] is initial position.
It is also helpful to know that velocity is related to acceleration as follows:

So let's do this problem in three pieces.
1 - A drag racer accelerates uniformly from rest straight track at 12m/s^2 for 5.5s.

This problem is nice because the units do not require any conversion (you are given acceleration in m/s^2 and time in seconds)
Let's find our speed at the end of 5.5 seconds

So at the end of the first leg, we have moved 181.5m and are zipping along at 60m/s
Now let's do the second leg as a 'whole new problem'.
maintains the speed it has reached for 3.0s
In this leg, the acceleration is 0 and the initial speed is where ended up at the end of the first leg - 60m.s
For this leg, we'll start our clock over again at 0

Now, we could have gotten more complicated and setup the equation to use to use the same clock as first leg, but that requires getting a lot more complex about where we are, what time it is, etc. More trouble than it is worth.
For now, just remember that we broke this problem into three small parts. And we will need to put humpty back together again in order to get a 'final answer'.
Ok, so we are at the end of the second leg. Let's see where we are now and see how fast we are going.
Our position is the sum of the first two legs. Our time is the sum of the first two leg's times.
Time at the end of the second leg is 5.5+3 = 8.5 seconds

Now we are ready for the last leg. At the end of 8.5 seconds, we are still going 60m/s since we did not change speeds on the second leg. We are 361 meters down the road.
Now we are told to -- slow down to a stop at the rate of -14m/s^2.
Again, let's treat this as a small standalone problem.
First, we need to figure out how long it takes to stop.

So it takes about 4.3 seconds to stop.

So add in the distance from the last leg to the total so far to get:
TotalDistance = 361.5 + 128.57
TotalDistance is approximately 490m

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