Auto & Truck Movements: Solving for Overtaking Distance & Speed

[PhilosophyofPhysics]

"At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s^2. At the same instant a truck, traveling with a constant speed of 9.5 m/s, overtakes and passes the automobile. (a). How far beyond the traffic signal will the automobile overtake the truck? (b). How fast will the automobile be traveling at that instant?"

 

[Doc Al]

Start by writing the kinematic formulas describing the distance of each as a function of time. One formula will involve accelerated motion; the other, constant speed motion.

 

[PhilosophyofPhysics]

hmm... I really think I have lost brain cells or something. In the back of the book it has the answers (a). 82m (b). 19 m/s

I get 82m if I use x=(2v^2)/a. But, I'm not sure how to derive that formula.

For the (b), I used v^2 = Vinitial^2 +2a(X-Xinitial) and got 19 m/s

 

[whozum]

The first equation is derived from the one you used in b.

 

[PhilosophyofPhysics]

The first equation is derived from the one you used in b.

oh wow, you're right. I never thought that a times x was just v^2. I was not thinking of the dimensions.

So after moving v initial over to the other side, I had delta V^2 = 2 v^2

Then i broke up delta V^2 into v times v. I change one of those into x/1 and 1/t. I combined v and 1/t to get a. So now I had a times x = 2v^2.

I get the first equation of my previous post after dividing by a on both sides.

x = (2v^2)/a


thanks, to both of you