Question

at the instant a traffic light turns green, a care starts with a constant acceleration of 6 ft per sec.At the same instant , a truck traveling with a constant velocity of 30 ft per sec passes the car. a) how far beyond its starting point will the car pass the truck?
b)how fast will the car be traveling when it passes the truck?

Answer 1

how long are the vehicles

Answer 2

it doesnt say:(

Answer 3

so we have to assume they are point particles/

Answer 4

yes ..i believe so

Answer 5

?

Answer 6

Im not sure

Answer 7

Acceleration of the car:
V(c) = 6t
Acceleration of the truck:
V(t) = 30

Integrate both to get velocity equations. Integrate both of those to get position equations. Set the position equations equal and solve for t.

Answer 8

Wait wait. Those are velocity equations. Integrate once.

Answer 9

so v(c) is 3t^2

Answer 10

and...v(t) is 30t

Answer 11

Velocity is the derivative of position, so
The integral of a velocity equation is a position equation.
You did the integrals correctly, but I would write them as:
P(c) = 3t^2
and
P(t) = 30t

Answer 12

Now, there's a middle step here usually. That is, I get a constant of integration.
P(c) = 3t^2 + C
but C is my starting position, and since you know that your starting position is 0, it drops out.
The same thing happens with the truck. If they started at different places, that step would be important.

Answer 13

Ohh..so if they started at different times, then we would use the C..and the C would just be when they started?

Answer 14

Right =D

Answer 15

Or rather, where they started.

Answer 16

ohhok! ...soo what would we do next?

Answer 17

Quick example, suppose they tell me that the truck starts 10 feet ahead of the car.
I'd get
P(T) = 30t + C
Well, I know when t = 0, the truck is at position 10, so that gives
30(0) + C = 10
so C = 10 and the position equation is
P(T) = 30t + 10

Answer 18

Well, we have position equations for both vehicles, and we're asked when they pass each other. They pass each other when their positions are the same, so set the position equations equal and solve for t. Let me know what time you get.

Answer 19

You should get two times, by the way. One of those times just isn't very interesting.

Answer 20

so is it 3t^2=30t?

Answer 21

Yes keep going.

Answer 22

1/10 t^2=t

Answer 23

what is the next step?

Answer 24

Been a while since you solved an equation like this?

Answer 25

It's easier if you gather everything on one side of the equation as your first step:
3t^2 = 30t
3t^2 - 30t = 0

Answer 26

so its 3t(t-10)=0 so t=0 or t=10?

Answer 27

Yes =D
And as a reminder, any time I have a quadratic equation, that is the highest exponent of the variable is 2, there are two good ways to solve it:
factor it out, this way sometimes works, but is often very quick.
use the quadratic formula. this way always works, but is a little slower.

Answer 28

So interpret those results. What does t=0 and t=10 mean? And why would I say that only one of those answers is interesting?

Answer 29

so, at 10 seconds...both the car and truck are at the same position.And t=10 is interesting..because t=0 is basically at 0 seconds

Answer 30

Yes. And my question already told me that they pass at that time. I didn't have to do any math to figure that out.

Answer 31

so the answer to b) is 60 ft/sec

Answer 32

and the answer to A)300 ft at 10 seconds?

Answer 33

Yes very good. Use the time you found in the velocity and position equations you wrote.

Answer 34

Thank you so much!

Answer 35

My pleasure =D