A driver going 40km/h breaks suddenly and stops his car over 10m. If he stops it from 80km/h with the same deceleration, what distance...?

I hate to point out the obvious but if the driver breaks at 40 km / hr he will be in hospital and won't be in a postion to try it again at 80 km/hr.

If, instead, he BRAKES from 40 km/hr then he would be a lot safer.
as the force of the brakes is the same at all speed
we have F * d = 1/2 m v^2
d = m v^2 / ( 2F)
everything is constant except for v and d so d is proportional to the SQUARE of v
double v and d is four times greater -> 40 m

When initial speed is 40 km/h :
vₒ = u m/s
v = 0 m/s
s = 10 m

v² = vₒ² + 2as
0 = u² + 2a(10)
20a = ‒u² ...... [1]

When initial speed is 80 km/h :
vₒ = 2u m/s
v = 0 m/s

v² = vₒ² + 2as
2as = ‒(2u)²
2as = ‒4u² ...... [2]

[2]/[1] :
s/10 = 4
Distance, s = 40 m


Alternative method :
F = ma
When deceleration is the same, the frictional force (F) is also the same.

Loss in kinetic energy = Work done by frictional force

When vₒ = 40 km/h = u m/s :
(1/2) m u² = F × 10 ...... [1]

When vo = 80 km/h = 2u m/s :
(1/2) m (2u)² = F × s ...... [2]


[2]/[1] :
2² = s/10
Distance, s = 40 m

"A driver going 40km/h breaks suddenly and stops his car over 10m."
You can use v^2 = vi^2 + 2ad to solve for a.

vi = initial velocity (given), v = final velocity (= 0), d = distance (given).

Now you know a.

"If he stops it from 80km/h with the same deceleration, what distance"
v^2 = vi^2 + 2ad, with a new value of vi.

You have for a constant braking deceleration :
-------------------------------------------------------------

dstop = ( v0 )^2 / ( 2 ) ( adec )

dstop = stopping distance in m
v0 = initial velocity in m/s
adec = stopping deceleration in m/s^2

Get adec from the 40 km/hr information :

v0 = ( 40 km/h ) ( 1000 m/km ) ( 1 h/ 3600 s ) = 11.11 m/s

adec = ( v0 )^2 / ( 2 ) ( dstop )

adec = ( 11.11 m/s )^2 / ( 2 ) ( 10 m ) = 6.173 m/s^2

Now apply this to the 80 km/h :

dstop = ( v0 )^2 / ( 2 ) ( adec )

dstop = ( 22.22 m/s )^2 / ( 2 ) ( 6.173 m/s^2 )

dstop = 40.0 m <---------

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W = Fd = mad = -1/2 mv0^2

d = - v0^2/(2a)

v0 is doubled while a remains the same --> d is quadrupled.

d = 4(10 m) = 40 m

d = V²/(2a); if V is doubled, d is quadrupled

d = 40 m

You either have a tire out of balance or snow has lodged itself onto one of the car's wheels causing an imbalance. If you make the car shudder based on speed then typically it's a wheel imbalance.

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