Finding the Equation for A Position vs. Time Graph from a Velocity vs. Time Graph

Problem: If a car is traveling at 65 kmph and the driver suddenly slams on the brakes and stops in 2.5 seconds, then what is the deceleration of the car from 65 kmph to 0 mph assuming that the car declines in speed linearly? How far did the car travel from when the driver slammed on his brakes until the car was at a complete stop?

This problem can best be approached by first constructing a graph.

· You also know that hours and seconds don’t agree because they are different units.

Then you can say that in one hour there are 3,600 seconds and there are 1,000 m in a km. So you can convert units like this:

65 km/h * h/3600 sec* 1000 m/km= about 18.1 m/sec

· Now you know that the car was going 18.1 m/sec and ended at 0 m/sec. And you know that the driver first touched his brakes at 0 sec and the car was completely stopped in 2.5 seconds.

· You can construct your graph where Velocity (y) vs. Clock Time (t). Plot your two points (0, 65) and (2.5, 0). And then construct a straight line between the two points because you know that the speed decreases linearly.

· You can next find the equation for the line by using y=at+b where y=velocity, t

=clock time, and a=acceleration (slope), b=y intercept (initial speed)

Slope=change in velocity/change in time

(18.1 m/sec – 0 m/sec)/(0 sec – 2.5 sec)= -7.24 m/s/s or –7.24 m/s^2

Therefore the car is accelerating at –7.24 m/s^2

· Now you can write your equation

y (velocity)=(-7.24m/s^2)x + 18.1 m/sec(initial speed and y intercept)

y=-7.24t + 18.1

· Next you need to find how far the car traveled in the 2.5 seconds. You can do this by converting the velocity vs. time graph into a position vs. time graph.

o The formula for changing a position vs. time graph to a velocity graph is

y=2at + b. Since a position vs. time graph has the equation y=at^2+b+c we can use the y=2at + b equation inversely.

o The values for y=2at + b for the velocity vs. clock time graph are a=-7.24, and b=18.1 and c=c

o Now you can work inversely by taking ½ of –7.24 and plugging that and 18.1 into y=at^2+b+c

o You get y=-3.62t^2 + 18.1t + c for you position vs. clock time equation

o Now that you can plug the time it took t=2.5 sec into y=-3.62t^2 + 18.1t + c

o y=-3.63(2.5)^2 + 18.1(2.5) + c

y=22.5625 + c

c=the initial starting position but that is irrelevant because all you need to know is how far it traveled.

`dy=c

y - `dy=(22.5625 + c) – (c)

y – ‘dy=22.5625

· Now you know that the car traveled 22.6 meters after applying the brakes at 65 kmph.