A bike with tires of radius 0.330 m speeds up from rest

to 5.33 m/s in 6.27's. Through

what angle do the wheels turn

in that time?

(Unit = rad)

The wheels turn 101.27 radians in that time.

Explanation:

First, we need to find the angular velocity of the tires. We use the following formula for this purpose:

Δv = rω

ω = Δv/r

where,

ω = Angular Velocity of Tires = ?

Δv = change in linear velocity = 5.33 m/s - 0 m/s = 5.33 m/s

r = radius of tire = 0.33 m

Therefore,

ω = (5.33 m/s)/(0.33 m)

ω = 16.15 rad/s

Now, the angular displacement covered by tires can be found out by using the general formula of angular velocity:

ω = θ/t

θ = ωt

where,

θ = angular displacement = ?

t = time = 6.27 s

Therefore,

θ = (16.15 rad/s)(6.27 s)

θ = 101.27 radians

higher potential

explanation:

because it is in a high speed so it must move into a of high potential