Too long to ask in title. Tricky Physics Question.?

Question

I've really being trying to figure this one out, if anyone could help me, I'd very much appreciate it.

If a car travelling at a speed of V can break at an emergency stop at a distance X. Assuming all other conditions are similar if the travelling speed of the car doubles, the stopping speed will be 4x. (I worked that part out)
A driver travelling at 40 km/h in a school zone can break to emergency stop in 3 meters. What would be the breaking distance of the car were travelling at 60 km/h.

Thanks in Advance.

Answer 1

Just like the 1'st question, the braking distance increases as the square of the ratio of the velocities. So
3*(60/40)² = 6.75 m

HTH

Doug

Answer 2

Assuming the brakes can apply a constant force and therefore constant deceleration, you can use the third equation of motion:

v^2 = u^2 + 2as

Now you want to stop so v=0. That means that the stopping distance is proportional to the square of the velocity.

Suppose s=3 when v=40. Then 1600 = -2a.
So if v=60, then 3600 = 2as = 1600s.
s = 3600/1600.

Please double check. My son's calling me and I have to go.

Answer 3

the potential energy of motion is calculated with

e = mv²

v is velocity. and as youll see it has a square, meaning that if you increase speed twofold, youll need 4fold the energy. if you increase speed 3 fold, you need 9 fold the energy.

i beleive you can calculate the rest yourself knowing this.

Answer 4

x/3 = (60/40)^2
x = 3*9/4 = 6.75 m.

Answer 5

i tried but failed sorry