Suppose that a 1000 kg car is traveling at 25 m/s. Its brakes can apply a force of 5000N. What is the minimum distance required for the car to stop?
2006-10-31 13:05:09 · 5 answers · asked by cubscaps33 5 in Science & Mathematics ➔ Physics
Force = Mass * Acceleration
5000 = (1000)a
a = 5(m/s)
Vfinal² = Vinitial² - 2ax
0 = 25² - 2(5)(x)
10x = 625
x = 62.5 m
2006-10-31 13:11:04 · answer #1 · answered by teh_popezorz 3 · 0⤊ 0⤋
If the comparable frictional forces prepare, the ensuing accelerations would be equivalent. via the placement-velocity relation, v^2=u^2 + 2as 0 = 15^2 + 2a*seventy six.5 a = -225/153 interior the 2d case, making use of the comparable equation, 0 = 32^2 + 2as a = -1024/2s right here, we've been given 2 equations for the comparable value a. Equating those values, -225 / 153 = -1024 / 2s s = 348.sixteen m for this reason the vehicle will stop after overlaying a distance of 348.sixteen metres.
2016-10-03 03:52:37 · answer #2 · answered by ? 4 · 0⤊ 0⤋
This is another way to "skin the cat":
Kinetic energy of car=Work done by applied force
1/2mv^2=F*s, where m =mass of car, v=velocity of car, F is the applied force, and s is the minimum required distance.
1/2*1000*25^2=5000*s
s=312500/5000
=62.5m
2006-11-01 03:42:32 · answer #3 · answered by tul b 3 · 0⤊ 0⤋
m=1000kg u=25m/s F=5000N
a=F/m
a=5000/1000=5m/s^2 a will be considered -ive as it is getting stopped
v^2=u^2+2as
s=25^2/(2*5)=62.5 m (ans)
2006-10-31 13:13:22 · answer #4 · answered by Anonymous · 0⤊ 0⤋
a=f/m=5000/1000=5 m/s^2
d=v^2/2a=625/10=62.5 m
2006-11-01 06:25:03 · answer #5 · answered by Anonymous · 0⤊ 0⤋