Physics question?

Question

If you were decelerating, but you doubled your initial speed...what does this do to the time and distance it takes to stop my car?

To do this would I just plug in random numbers?

I still don't get why you would double the time.

Answer 1

Summarizing the information that you have given:

A car begins to decelerate to a complete stop, we want to know that if it is moving at a a velocity that's twice the initial velocity, how does that affect the time and distance it takes to stop the car.

So we want to compare the difference between v(initial)=v(0)
or v(initial)=2v(0) (where v(0) is some constant velocity) when we start to decelerate.

Assuming that the initial velocity is constant, and that the deceleration is also a constant, then:

x(1) = x(0) + v(0)t -(1/2)at^2, this is v(initial)=v(0)

x(2) = x(0) + 2v(0)t -(1/2)at^2, this is v(initial)=2v(0)

To see the difference, we substract x(1) from x(2):

x(2) - x(1) = v(0)t

This means that the distance increased by v(0)t if we doubled the velocity to 2v(0).

Now v(final) = v(initial) - at implies that t = v(initial)/a

So it's obvious that if we double the initial velocity, we also double the amount of time that it takes the car to come to a complete stop. And plugin t into x(2) - x(1), we see that

x(2) - x(1) = v(initial)t = v(0)^2/a for v(initial) = v(0)

x(2) - x(1) = v(initial)t = 4v(0)^2/a for v(initial) = 2v(0)

So if we have doubled the time, then we quadruple the distance it takes to stop the car.

Answer 2

The time it takes to stop is v0 - at. if you double V0, t must double. The distance it takes is s = .5*a*t^2. Double V0, double t; quadruple s.

Answer 3

doubled. use random numbers.

Answer 4

more distance