How much ahead the front axle does the wheel jet the water that it pick

[mmoadi]

Homework Statement

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A bicycle with a speed 12 km/h is driving along a leveled wet road. How much ahead the front axle (see picture) does the wheel jet the water that it picks up from the floor?The radius of the wheel is r = 35 cm. Let’s suppose that the drops of water that are separating from the highest point of the wheel are flying the farthest; and we suppose that the drops are not separating from the lower part of the wheel because of the mudguard of the wheel is preventing them. What is the solution for the bicycle without the mudguard under the same condition?

Homework Equations

∆y= y- y0= v(0)*t- ½ gt²
∆x= x- x(0)= v(0)t

The Attempt at a Solution

First I calculated time:
∆y= y- y(0)= v(0y)*t- ½ gt²
y0= 0
v(0y)= 0
and I set for y= (-0.7m), because the drop is falling to the ground

y= -½ gt²
t= 0.38 s

Now I calculated the distance:
∆x= x- x0= v(0x)*t
x0= 0
v(0x)= 12 m/s
t= 0.38 s

x= v(0x)*t
x= 1.27 m

_{Are my attempts to solve this problem correct?
Thank you!}

 

[anathema]

Your attempts to solve the problem are partially correct. You correctly calculated the time it takes for the drop of water to fall from the highest point of the wheel to the ground. However, your calculation for the distance traveled by the drop is incorrect. The equation ∆x = v(0x)*t only applies when the acceleration is constant, which is not the case here. In this situation, the drop is accelerating due to gravity, so you cannot use this equation.

To calculate the distance traveled by the drop, you can use the equation ∆y = ½ gt², which you already used to calculate the time. However, you need to use the final velocity of the drop, which is 0 m/s, since it comes to a stop at the ground. So the correct calculation for the distance traveled by the drop is:

∆y = ½ gt²
∆y = ½ * 9.8 m/s² * (0.38 s)²
∆y = 0.68 m

Therefore, the drop of water travels 0.68 meters ahead of the front axle of the bicycle.

For the second part of the problem, you can use the same equation to calculate the distance traveled by the drop without the mudguard. However, in this case, the drop can separate from the lower part of the wheel as well, so the calculation will be slightly different. You will need to take into account the distance traveled by the drop from the lower part of the wheel to the ground, in addition to the distance traveled from the highest point of the wheel to the ground. The calculation will be:

∆y = ½ gt² + ∆y'
Where ∆y' is the distance traveled by the drop from the lower part of the wheel to the ground. This distance can be calculated using the equation ∆y' = ½ gt², where t is the time it takes for the drop to reach the ground from the lower part of the wheel. This time can be calculated using the same method as before. So the final calculation for ∆y' is:

∆y' = ½ * 9.8 m/s² * (0.38 s)²
∆y' = 0.68 m

Therefore, the total distance traveled by the drop without the mudguard is:

∆y = ½ gt² + ∆y'
∆y

 

[kerimek]

Your attempts to solve this problem are correct. The distance the drop of water travels horizontally is equal to the velocity of the bicycle (12 km/h or 3.33 m/s) multiplied by the time (0.38 s), which results in a distance of 1.27 m. This means that the front axle of the bicycle is 1.27 m ahead of the point where the drop of water falls to the ground.

Without the mudguard, the drops of water would not be prevented from separating from the lower part of the wheel, which means they would travel a farther distance. In this case, the solution would be the same except for the fact that the drop of water would separate from the lower part of the wheel, resulting in a longer distance traveled.

It is important to note that this solution is an approximation and may vary depending on the specific conditions of the road and the bicycle. Additionally, the angle at which the drop of water separates from the wheel may also affect the distance traveled.