Can someone help me?

Physics

1 answer:

Leviafan [203]3 years ago

4 0

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

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a_sh-v [17]

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PEi + KEi = PEf + KEf

Since the object is falling from rest, the initial kinetic energy is zero. Also, since the object hits the ground at its final position, the final potential energy is zero. This leaves:

PEi = KEf
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*cancel out mass on both sides of the equation

gh = 0.5v^2
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