Abicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф=t

Ф= rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m

The bicycle traveled about 22 m during the 5.0 seconds of braking

Centripetal Acceleration can be formulated as follows:

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Centripetal Force can be formulated as follows:

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

Grade: High School

Subject: Physics

Chapter: Circular Motion

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s