Distance Traveled by a Car Before Stopping

[Erin_Sharpe]

Ok... this is a long one, please bear with me!

A car is driving at a constant velocity of 18 m/s. There is a school on the road with the stop sign extended. The car is 40 m away from bus when he sees the sign. There is a time dealy of 0.75 seconds bewtween the time the driver sees the sign ans when the driver can begin to slow down. During the reaction time the distance "d" in meters traveled by the car is given by the equation d= 18t, "t" is time in seconds from when the driver sees the bus.
When the brakes are applied after the 0.75 second reaction time, the equation is: d= -3t^2 + 22.5t - 1.6875.
After the brakes are applied it takes 3 seconds for the car to come to a stop. These 3 seconds plus the 0.75 second driver reaction time means the care stops 3.75 seconds after seeing the school bus.

What is the piecewise-defined function to describe the distance traveled by the car until it stops
How far does the car travel before stopping?
Does the car hit the bus?


I'd appreciate some tips on this one!

Thanks guys!
Erin

 

[HallsofIvy]

Your title "piecewise linear functions" is misleading. For the first 0.75 seconds, the function is linear, in fact it is exactly the d= 18t you are given, but after that it is quadratic.

All you need to do is "patch" those two formulas together. The first "piece" is simply 18t for 0< t< 0.75. The second piece is the second formula: for 0.75< t< 3,
d= -3t²+ 22.5t- 1.6875. Write those in the standard for for "piecewise functions".

 

[gerben]

I wonder why the speed during the reaction time (18) is smaller than the speed when starting to apply the brakes (22.5).
Does it have any relation with this thread:
https://www.physicsforums.com/showthread.php?t=60150&page=1&pp=15

 

[HallsofIvy]

I wonder why the speed during the reaction time (18) is smaller than the speed when starting to apply the brakes (22.5).

The speed "when starting to apply the brakes" is not 22.5.

The speed according to the second formula is given by -6t+ 22.5.
At t= 0.75, that is -6(3/4)+ 22.5= -9/2+ 45/2= 36/2= 18 m/s.
That is, of course, the speed given by the first formula.

 

[gerben]

Ah, I see

but what an unpractical way to express the distance traveled during braking, you will have to subtract -3(0.75)² + 22.5*0.75 - 1.6875 from the equation to know how much distance is traveled during braking

 

[dumguy]

it takes 40.5 m for the car to stop and yes it hits the bus. If you attempt to use the formulas individually it does not work out. If you input the entire piecewise into your TI-83 it should give the right answer.