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.

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

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

so above is the instantaneous accelerations