Question

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

a.what is the average velocity of the particle between t = 0 and t = 2.0 s?
b.what is the instantaneous velocity of the particle at t = 0, 1.0, and 2.0 s?
c.what is the average acceleration of the particle between t = 0 and t = 2.0 s?
d.what is the instantaneous acceleration of the particle at t = 0, 1.0, and 2.0 s?

Answer 1

Part a)

At t = 0 the position of the object is given as

`x = 0`

At t = 2

`x = 2 sin(\pi/2) = 2cm`

so displacement of the object is given as

`d = 2 - 0 = 2cm`

so average speed is given as

`v_(avg) = (2)/(2) = 1 cm/s`

Part b)

instantaneous speed is given by

`v = (dy)/(dt)`

`v = 2cos(\pi t/4 ) * (\pi)/(4)`

now at t= 0

`v = (\pi)/(2) cm/s`

at t = 1

`v = 2 cos(\pi/4) * (\pi)/(4)`

`v = (\pi)/(2\sqrt2)`

at t = 2

`v = 0`

Part c)

Average acceleration is given as

`a_(avg) = (v_f - v_i)/(t)`

`a_(avg) = (0 - (\pi)/(2))/(2)`

`a = -(\pi)/(4) cm/s^2`

Part d)

Now for instantaneous acceleration

As we know that

`a =- \omega^2 y`

at t = 0

`a = -(\pi^2)/(16) * 0 = 0 cm/s^2`

at t = 1

`y = \sqrt2 cm`

now we have

`a = -(\pi^2)/(16)*\sqrt2`

At t = 2 we have

`y = 2 cm`

`a = -(\pi^2)/(16)*2`

`a = -(\pi^2)/(8)`

so above is the instantaneous accelerations