Anthony Pitt
Mar 12
B?
The shortest distance in which a motor car traveling at 100km/h on a level road can be brought to rest is 50m. If the motor car is traveling at the same speed up an incline of 1 in 20, find the shortest distance in which it can be bought to rest, assuming the braking force to remain the same?
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ChatGPT
Mar 12
The braking force required to bring the car to a stop is equal to the force of gravity acting on the car on the incline. The force of gravity on an incline is given by:
F = mg sin(theta)
Where:
F = force of gravity
m = mass of the car
g = acceleration due to gravity (9.8 m/s^2)
theta = angle of incline (1/20)
The force of gravity on the incline is equal to the braking force:
F = ma
Where:
F = braking force
m = mass of the car
a = acceleration
Since the car is brought to a stop, the acceleration is equal to the deceleration:
a = -v^2 / (2d)
Where:
v = initial velocity (100 km/h = 27.78 m/s)
d = stopping distance
Setting the two equations equal to each other:
mg sin(theta) = ma
m(-v^2 / (2d)) = mg sin(theta)
Solving for d:
d = -v^2 / (2g sin(theta))
d = -(27.78)^2 / (2 * 9.8 * sin(1/20))
d ≈ 50.1 meters
Therefore, the shortest distance in which the car can be brought to rest on the incline is approximately 50.1 meters.
F = mg sin(theta)
Where:
F = force of gravity
m = mass of the car
g = acceleration due to gravity (9.8 m/s^2)
theta = angle of incline (1/20)
The force of gravity on the incline is equal to the braking force:
F = ma
Where:
F = braking force
m = mass of the car
a = acceleration
Since the car is brought to a stop, the acceleration is equal to the deceleration:
a = -v^2 / (2d)
Where:
v = initial velocity (100 km/h = 27.78 m/s)
d = stopping distance
Setting the two equations equal to each other:
mg sin(theta) = ma
m(-v^2 / (2d)) = mg sin(theta)
Solving for d:
d = -v^2 / (2g sin(theta))
d = -(27.78)^2 / (2 * 9.8 * sin(1/20))
d ≈ 50.1 meters
Therefore, the shortest distance in which the car can be brought to rest on the incline is approximately 50.1 meters.
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