A motorcyclist is traveling at 51.9 mph on a flat stretch of highway during a sudden rainstorm. The rain has reduced the coeffic
ient of static friction between the motorcycle's tires and the road to 0.076 when the motorcyclist slams on the brakes. (a) What is the minimum distance required to bring the motorcycle to a complete stop? (b) What would be the stopping distance if it were not raining and the coefficient of static friction were 0.580?
2 answers:
Answer: a) 361.23m
b) 47.38m
Explanation:
Friction is opposite in direction to an applied force
Find the attached file for the solution
(b) The distance traveled with the friction of rain-free conditions is 47.38 m.
Answer:
Part (a) 361m
Part (b) 47.4m
This problem involves involves the nonconservative frictional force. To solve this problem. The energy approacb will be used. The energy conservation is used to relate the work done by the frictional force over the required distance to the kinetic energy of the motorcycle before the brakes are applied.
Explanation:
The equation is
K1 + U1 + Wf = K2 + U2
Where U standa for potential energy which in this case is zero as the height of the motorcycle does not change over the distance of the application of the brakes to the tyres. The work of friction is negative.
The full solution can be found in the attachment below.
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The Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s
<h3>How to determine the inertia</h3>
Using the formula:
I = 1/2 M₁R₁² + 1/2 M₂R₂²
Where I = Inertia
I = 1/2 * 0.810* (2. 60)² + 1/2 * 1. 58 * (5)²
I = 1/2 * 5. 476 + 1/2 * 39. 5
I = 2. 738 + 19. 75
I = 22. 488 kg. m²
To determine the block's speed, use the formula
v =
v =
v =
v = 6. 4 m/s
Therefore, the Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s
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Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.