distance required to stop the car?

Question

You are driving your car on a straight level road and slam on your brakes and skid to a stop over a distance of 20m(your car doesn't have anti-lock brakes). If you had been traveling twice as fast, what distance would be required to stop your car under the same conditions?

a) 28m
b) 30m
c) 40m
d) 60m
e) none of the above

Answer 1

This is a Kinetic Energy problem. The equation for KE is:

KE = (1/2)(mass)(velocity)^2

So, when you double the velocity, the KE increases four times (the velocity is squared, 2x2 = 4). Thus, the car's stopping distance increases 4 times (because the kinetic energy has increased 4 times).

Therefore, the stopping distance at double the speed is 4 x 20m = 80m. Thus, the answer is "e" - none of the above.

Best wishes and good luck.

Answer 2

The brakes exert the same braking force regardless of speed (friction force is proportional to weight). If the force is the same, the acceleration is the same. The applicable equation is

v^2 = 2*a*d

Since the velocity is a squared term, it will take 80m to stop the card. The answer is (e).

Answer 3

As is well known,
x(t) = x(0) + v(0)t + at^2/2
(In this case, a is negative.)

In this case, x(0) = 0
v(t) = dx/dt = v(0) + at, so v(t) = 0 when t = -v(0)/a
and then x(-v(0)/a) = 0 + v(0)(-v(0)/a) + a(v(0)/a)^2/2
= -(v(0)^2/a) + (v(0)^2/2a) = -(v(0)^2/2a)

So if v(0) were twice as large, x(-v(0)/a) would be 4 times as large. So 20 meters would become 80 meters.

So the right answer is e).

Answer 4

is that from a 1936 test paper modern cars can stop much quicker than that, there is no correlation between stopping from 36 mph and 71 mph, the energy that needs to be dissipated at 71 mph is non linear

Answer 5

the formula is this;

x=(v^2)/2g

so when the v is twice the brake length is 4 times more
so it would be ; 4*20=80

Answer 6

Not enough info..need coefficient of friction of the road..mass of car...velocity of car...can't assume that the answ is 2 x 20m = 40m, cause might not be linear.

Answer 7

♠ u were traveling with speed v1=a*t1; your baking distance d1=0.5a*t1^2, where d1=20m, t1 time required to stop, a is deceleration;
♣ u are traveling with speed v2=a*t2=2*v1; your baking distance d2=0.5a*t2^2;
♦ thus d2/d1 =(t2/t1)^2, and v2/v1 = t2/t1, hence d2/d1= (v2/v1)^2=
= d2/20 =2^2, hence d2=80m;