Question
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?
Answers
bobpursley
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.
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.
Barndon
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
bobpursley
No, I cant. consider your third equation, isn't it what I used?
Brandon
please can u answer me i am stressing out
Brandon
ohhhhhhhhhhh i get it now thanks