Respuesta :
Part a)
At t = 0 the position of the object is given as
[tex] x = 0[/tex]
At t = 2
[tex] x = 2 sin(\pi/2) = 2cm[/tex]
so displacement of the object is given as
[tex]d = 2 - 0 = 2cm[/tex]
so average speed is given as
[tex]v_{avg} = \frac{2}{2} = 1 cm/s[/tex]
Part b)
instantaneous speed is given by
[tex] v = \frac{dy}{dt}[/tex]
[tex]v = 2cos(\pi t/4 ) * \frac{\pi}{4}[/tex]
now at t= 0
[tex]v = \frac{\pi}{2} cm/s[/tex]
at t = 1
[tex] v = 2 cos(\pi/4) * \frac{\pi}{4}[/tex]
[tex]v = \frac{\pi}{2\sqrt2}[/tex]
at t = 2
[tex]v = 0[/tex]
Part c)
Average acceleration is given as
[tex]a_{avg} = \frac{v_f - v_i}{t}[/tex]
[tex]a_{avg} = \frac{0 - \frac{\pi}{2}}{2}[/tex]
[tex]a = -\frac{\pi}{4} cm/s^2[/tex]
Part d)
Now for instantaneous acceleration
As we know that
[tex]a =- \omega^2 y[/tex]
at t = 0
[tex] a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2[/tex]
at t = 1
[tex]y = \sqrt2 cm[/tex]
now we have
[tex]a = -\frac{\pi^2}{16}*\sqrt2[/tex]
At t = 2 we have
[tex] y = 2 cm[/tex]
[tex]a = -\frac{\pi^2}{16}*2[/tex]
[tex]a = -\frac{\pi^2}{8}[/tex]
so above is the instantaneous accelerations
