PHysics question about braking distance?

Question

If the speed of a car is increased by 65%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time.

Answer 1

D = V^2/2a
So D2/D1 = (V2/V1)^2 = 1.65^2 = 2.72, or by 172%.

Answer 2

The equations for constant acceleration are:

Vf = Vi + A x T
D = Vi x T + (1/2) x A x T^2

where
A is the acceleration
T is the time interval
Vi is the initial velocity
D is the distance traveled

In this case, we are interested in braking so Vf is 0 and A is the maximum acceleration the brakes, tires, and road can provide.

The velocity equation gives us the braking time from a velocity of Vi and the distance equation uses that T to give us the minimum braking distance.

let's add in a factor k for the initial velocity. That gives us

T(k) = k x (-Vi) / A = k x Ti

Multiplying the initial velocity by a factor of k multiplies the time to stop by a factor of k.

Substituting in th equation for distance we get:

D = (k x Vi) x (k x Ti) + (1/2) x A x (k x Ti)^2

Both terms on the right have a factor of k^2 so we get

D = (k^2) x ((Vi x Ti) + (1/2) x A x Ti^2)

so D goes up by a factor of k^2. Doubling the initial velocity causes the breaking distance to quadruple.

Answer 3

1.65^2