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...?
回答 (8)
2015-08-29 8:22 am
✔ 最佳答案
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
2015-08-28 9:13 pm
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
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
2015-08-28 8:50 pm
"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 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.
2015-08-29 7:19 am
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 <---------
Please award Best Answer to the most correct and most helpful response to your question.
-------------------------------------------------------------
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 <---------
Please award Best Answer to the most correct and most helpful response to your question.
2015-08-29 1:31 am
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 = - v0^2/(2a)
v0 is doubled while a remains the same --> d is quadrupled.
d = 4(10 m) = 40 m
2015-08-28 10:30 pm
d = V²/(2a); if V is doubled, d is quadrupled
d = 40 m
d = 40 m
參考: Conservation of energy
2016-03-02 6:59 am
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.
2015-08-30 5:08 am
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