Posted by **Brandon** on Wednesday, September 17, 2008 at 3:58pm.

As a traffic light turns green, a waiting car starts with a constant acceleration of 6.0 m/second squared. At the instant the car begins to accelerate, a truck with the constant velocity of 21 m/ second squared passes in the next lane. HINT:Set the 2 distance equations equal to each other. How far will the car travel before it overtakes the truck? How fast will the car be traveling when it overtakes the truck?

- Physics -
**bobpursley**, Wednesday, September 17, 2008 at 4:07pm
The "Hint" is the solution.

distancecar1=1/2 a*t^2

distancecar2=vo*t + 1/2 a t^2

but car2 has a=0

set distances equal, so

1/2 at^2=vot and solve for t.

- Physics to bobursley -
**Barndon**, Wednesday, September 17, 2008 at 4:56pm
can u go back to my problem and explain it differently because my teacher has only taught us 4 equations of motion and they r d=(v intial +v final)/2 times t, v final=v intial+at, d= v intial(t)+1/2at^2, v final^2=v intial^s + 2ad

- Physics -
**bobpursley**, Wednesday, September 17, 2008 at 4:59pm
No, I cant. consider your third equation, isn't it what I used?

- Physics for bobpursley -
**Brandon**, Wednesday, September 17, 2008 at 5:07pm
ohhhhhhhhhhh i get it now thanks

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