posted by Brandon .
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?
The "Hint" is the solution.
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.
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
please can u answer me i am stressing out
No, I cant. consider your third equation, isn't it what I used?
ohhhhhhhhhhh i get it now thanks