A car is moving 15 m/s skids to a stop in 20m. If the car travels at 45 m/s. How far will it skid, assuming .. same constant braking force?
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The work done on the car (by the friction in the road) equals the loss of kinetic energy in the car.
work done on car = loss of KE
force × distance = ½(mass)v²
From that you can see that the distance is proportional to the square of the speed; so if the speed gets doubled, the braking distance gets quadrupled; etc.
If you still need help: apply that equation to the first case (the first skid):
F × (20meters) = ½(mass)(15meters/s)²
Then apply it to the second case, using the variable "d" to stand for the unknown distance:
F × (d) = ½(mass)(45meters/s)²
Now you can use algebra to combine those two equations and solve for "d"
If you STILL need help: Divide the second equation by the first equation. That gives you:
d / (20meters) = (45meters/s)² / (15meters/s)²
Multiply both sides by "20meters" to get:
d = (20meters) × (45meters/s)² / (15meters/s)²
work done on car = loss of KE
force × distance = ½(mass)v²
From that you can see that the distance is proportional to the square of the speed; so if the speed gets doubled, the braking distance gets quadrupled; etc.
If you still need help: apply that equation to the first case (the first skid):
F × (20meters) = ½(mass)(15meters/s)²
Then apply it to the second case, using the variable "d" to stand for the unknown distance:
F × (d) = ½(mass)(45meters/s)²
Now you can use algebra to combine those two equations and solve for "d"
If you STILL need help: Divide the second equation by the first equation. That gives you:
d / (20meters) = (45meters/s)² / (15meters/s)²
Multiply both sides by "20meters" to get:
d = (20meters) × (45meters/s)² / (15meters/s)²
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W = FΔD = ΔKE
(-F)ΔD = 1/2 m(vf)^2 - 1/2 m(vi)^2
Note: The breaking force is opposite the direction of motion so there is a -F. Also, since the final velocity is zero (the car comes to a stop), 1/2 m(vf)^2 = 0
-maΔD = - 1/2 m(vi)^2
aΔD = 1/2 (vi)^2
a = (vi)^2 / (2ΔD)
a = (15 m/s)^2 / (2*20 m) = 5.625 m/s/s
Since the same breaking force is applied in the second scenario, the acceleration (or deceleration for that matter) is the same.
ΔD = (vi)^2 / (2a)
ΔD = (45 m/s)^2 / (2*5.625 m/s/s) = 180 m
The car skids 180 m
(-F)ΔD = 1/2 m(vf)^2 - 1/2 m(vi)^2
Note: The breaking force is opposite the direction of motion so there is a -F. Also, since the final velocity is zero (the car comes to a stop), 1/2 m(vf)^2 = 0
-maΔD = - 1/2 m(vi)^2
aΔD = 1/2 (vi)^2
a = (vi)^2 / (2ΔD)
a = (15 m/s)^2 / (2*20 m) = 5.625 m/s/s
Since the same breaking force is applied in the second scenario, the acceleration (or deceleration for that matter) is the same.
ΔD = (vi)^2 / (2a)
ΔD = (45 m/s)^2 / (2*5.625 m/s/s) = 180 m
The car skids 180 m
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Kinetic Energy = 0.5Mv^2 = Work = Force * distance
M is constant ===> Energy is a function of v^2 =====> We tripled v ===> Energy increased by 3^2 = 9 times the energy at 45 m/s
Work was increased by 9 times
Force is constant ===> Distance must increase by 9 times
D = 9 * 20 = 180 m
M is constant ===> Energy is a function of v^2 =====> We tripled v ===> Energy increased by 3^2 = 9 times the energy at 45 m/s
Work was increased by 9 times
Force is constant ===> Distance must increase by 9 times
D = 9 * 20 = 180 m