Physics question?

Question

A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It travels a distance d before coming to rest. If its initial velocity were doubled, the distance required to stop would? A) quadruple B) Stay the same C) double as well D) decrease by a factor of two please explain why also

Answer 1

the formula is
d = (v'^2 - v^2) / 2*a
now v' = 0 and a is negative since it is decelerating...
simplifying
d = v^2 / 2*a

if the velocity, v is doubled, the new distance d' is
d' = (2*v)^2 / 2*a
d' = 4*v^2 / 2*a
since the original distance, d = v^2/2*a...
thus,
d' = 4*d
thus, the new distance would be four times the original distance.

Answer 2

Well, the TIME to come to rest would double. But since half the new time would be spent just slowing down to your original velocity, and would cover much more distance in that time (due to the higher speed), the total braking distance would approximately quadruple.

So, A) quadruple.

Answer 3

The distance traveled under constant deceleration is given by

d = v0*t - 0.5*a*t^2

The car stops when a*t = v0, so t = v0/a substituting that back gives

d = v0*^2/a - 0.5*v0^2/a

d= 0.5*v0^2/a

Since d is proportional to the square of v0, doubling v0 will quadruple d

Answer 4

i think it would be c, double as well.

if your v and a are constant, and it takes disatnce d to rest, then if your v=2v and a is the same, then your distance would be 2d to rest.

your original equation is:

v-a=d

new equation:

2v-a=xd

since v is doubled, and your a is not, it will take twice the distance it did before to have the object rest

Answer 5

A.