An object moving at a constant speed requires 4.0 s to go once around a circle with a diameter of 5.0 m. what is the magnitude of the instantaneous acceleration of the particle during this time? o2.20 m/s2 3.93 m/s2 6.17 m/s2 12.3m/2 15.4 m/g2

Acceleration, a = 6.17 m/s²

Explanation:

It is given that,

Diameter of the circular track, d = 5 m

Radius of circular track, r = 2.5 m

An object moving at a constant speed requires 4.0 s to go once around a circle. We need to find the instantaneous acceleration of the particle during this time. It is given by :

Where

= angular velocity

or

So, the  instantaneous acceleration of the particle during this time is 6.17 m/s². Hence, this is the required solution.

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

inverse v=i*r

nice nice