if the speed of the car increasesby 50%,by what factor does this increase the stopping distance?

Answer 1

surely, no
If you increase the speed, it will take longer distance to stop

Answer 2

Stopping distance (that is - braking distance as opposed to thinking distance) is proportional to the square of the speed. So if speed increases by 50% (a factor of 1.5), stopping distance increases by a factor of 1.5 squared = 2.25, i.e. 125% increase.

Answer 3

Suppose the car is moving on an equipotential surface.

let the kinetic energy be "E".
If this body is put to rest,then according to conservation of mech.energy E=k*x1 {k=break constant, x =displacement}
"50%" increase implies that ,now kinetic energy=1.5*E.
So stopping dist, x2=1.5*E/k.
x2-x1=E/2k.
so,percentage increase={(E/2k)/(E/k)}*100="50% increase"

Answer 4

u2 = 1.5 u1

(S2/S1) = (u2/u1) ^2 = (1.5)^2 = 2.25

S2 = 2.25*S1

Stopping distance increases by 2.25 times for the same deceleration.