Phineas is driving a car down the street at a certin speed, he slams on the brakes, causing the car to immediately go into a skid. He comes to a stop 100 ft later. Ferb takes a turn driving the same but at a speed twice as fast as Phineas' speed. He slams on the brakes and immediately begins to skid. How far does he travel before he comes to a stop?

a)50 ft.

b) 100 ft.

c) 200 ft.

d) 400 ft.

e) 10,000 ft

f)Impossible to tell, more information is needed

d. 400 ft.

Step-by-step explanation

The energy of the car when the Phineas was driving is equal to KE = ½·m·v², where
v is the speed at which Phineas was driving,
and m is the mass of the car (mass of the drivers are negligible)

The breaking force is constant for in both drivers (as nothing in the tires, or the road conditions change). Let's denote this force by "F".

Then, the work done by the friction will be equal to the entire kinetic energy of the car (so as to stop it). Also recall that work is equal to force multiplied by the parallel distance. (W = F·d). Thus,

F·(100 ft) = ½·m·v² = KE

When Ferb drives, he drives with twice the speed. Thus, his kinetic energy will be : ½·m·(2v)² = 4·[½·m·v²] = 4·KE

Thus, we have 
F·(100 ft) = KE
F·(Ferb's stopping distance) = 4·KE

This means that Ferb's stopping distance should be 400 feet