Physics Kinematic

Question

Wile E. Coyote has strapped himself to an ACME rocket and is moving along at 25.0 m/s in pursuit of the roadrunner. As he is cruising along he realizes that he is heading directly for the edge of a cliff. He drags his feet in order to slow down at a constant rate of 5.86 m/s^2 until he comes to rest. If the cliff is 36.0 meters away, by how much does he overshoot the edge?

Answer 1

Hi!

I hope this will help you to understand the way to solve the question.

First, we know this is a kinematic problem so we need to determine which SUVAT equation we will use.

So let's go through each letter.

S: We don't know the distance but this is what we are trying to find out. I like to note this with a question mark.

So so far we have s=?

Now U, the initial velocity, we are told is 25m/s.

So we have s=? u=25

V, well Wile E Coyote is trying to stop so V is 0.

s=? u=25 v=0

A, we are told that the dragging of the feat causes a deceleration of 5.86, in terms of acceleration, is it -5.86m/s/s

Now we have, s=? u=25 v=0 a=-5.86

T is all that is left but time is not mentioned and we don't need to us it.

We have to find the equation that incorporates all the letters we have. That is v²=u²+2as

So we rearrange for what we're trying to find, s.

This gives us (v²-u²)/2a = s

Sub in our values.

(0²-25²)/2(-5.86) = s

s= 53.33m

But we don't just stop there. We have to find out how much he overshoots by.

We are given the distance to the edge of the cliff so we subtract that from the total distance to give us 17.33.