The wheels of a skateboard roll without slipping as it accelerates at 0.45 m/s2 down an 75-m-long hill.

If the skateboarder travels at 1.8 m/s at the top of the hill, what is the average angular speed of the 2.6-cm-radius whe els during the entire trip down the hill?

I first found out the velocity of the skater which was 38.4 m/s using v^2=vo^2+2aX and then plugged the velocity into v=vo+at to find time but the time came out to be 3.73 s which is too little? I'm not sure if what I did is right.. I was going to plug the time into w=wo+at to find the angular velocity.

To find the average angular speed of the skateboard wheels during the entire trip down the hill, we need to first calculate the final velocity of the skateboarder at the bottom of the hill.

We can use the equation v^2 = vo^2 + 2ax, where v is the final velocity (1.8 m/s), vo is the initial velocity (0), a is the acceleration (0.45 m/s^2), and x is the distance traveled (75 m).

Using this equation, we can rearrange it to solve for the final velocity v:

v^2 = 0^2 + 2(0.45 m/s^2)(75 m)
v^2 = 0 + 67.5 m/s^2 * m
v^2 = 67.5 m^2/s^2
v = √(67.5 m^2/s^2)
v ≈ 8.2 m/s

Now, to find the time it takes for the skateboarder to travel down the hill, we can use the equation v = vo + at, where v is the final velocity (8.2 m/s), vo is the initial velocity (1.8 m/s), a is the acceleration (0.45 m/s^2), and t is the time.

Using this equation, we can rearrange it to solve for time t:

8.2 m/s = 1.8 m/s + (0.45 m/s^2)t
8.2 m/s - 1.8 m/s = (0.45 m/s^2)t
6.4 m/s = (0.45 m/s^2)t
t = 6.4 m/s / 0.45 m/s^2
t ≈ 14.22 s

Now, to find the average angular speed, we can use the equation w = Δθ / t, where w is the average angular speed, Δθ is the change in angle, and t is the time.

The change in angle can be found using the formula Δθ = Δs / r, where Δs is the distance traveled (75 m) and r is the radius of the wheels (2.6 cm = 0.026 m).

Δθ = 75 m / 0.026 m ≈ 2884.62 radians

Using this value for Δθ and the value of t that we found earlier, we can calculate the average angular speed:

w = Δθ / t
w = 2884.62 radians / 14.22 s
w ≈ 202.64 radians/s

Therefore, the average angular speed of the 2.6-cm-radius wheels during the entire trip down the hill is approximately 202.64 radians/s.

To find the average angular speed of the wheels during the entire trip down the hill, you need to first calculate the final angular speed of the wheels at the bottom of the hill.

To do this, you can use the equation:

v = ω * r

Where:
v is the linear velocity (1.8 m/s in this case),
ω (omega) is the angular velocity you are trying to find, and
r is the radius of the wheels (2.6 cm or 0.026 m).

Rearranging the equation, you get:

ω = v / r

Plug in the values:

ω = 1.8 m/s / 0.026 m = 69.23 rad/s

Now that you have the final angular speed, you can calculate the average angular speed using the equation:

ω_avg = (ω_initial + ω_final) / 2

In this case, the skateboard starts from rest at the top of the hill, so the initial angular speed is zero. Plug in the values:

ω_avg = (0 + 69.23 rad/s) / 2 = 34.62 rad/s

Therefore, the average angular speed of the wheels during the entire trip down the hill is 34.62 rad/s.

Did you get the answer?

Somebody once told me the world is gonna roll me