The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion

Question

s = 2 sin(πt) + 2 cos(πt),
where t is measured in seconds. (Round your answers to two decimal places.)
(a) Find the average velocity during each time period.

Answer 1

(a) you give no time periods, but the average velocity over [a,b] is just the total distance divided by the total time:
(s(b)-s(a))/(b-a)

Answer 2

(i) [1, 2]
cm/s

(ii) [1, 1.1]
cm/s

(iii) [1, 1.01]
cm/s

(iv) [1, 1.001]
cm/s

(b) Estimate the instantaneous velocity of the particle when t = 1.
cm/s

Answer 3

oobleck told you how to do the average
(position at end - position at beginning) / time elapsed

for instantaneous, differentiate
ds/dt = velocity = d/dt[ 2 sin(πt) + 2 cos(πt) ]
v(t) = 2 pi cos ( pi t ) - 2 pi sin (pi t)
if t = 1
v = 2 pi cos pi - 2 pi sin pi
= 2 pi (-1) - 0
= -2 pi

Answer 4

By the way they luckily asked for displacement and velocity, not average distance and speed. If they asked for the scalars and not the vectors you would have to be careful about going back and forth. If you go all the way around a circular track your displacement and average velocity is ZERO. However your distance is pi D and speed is pi D/time

Answer 5

This problem is trying to get you to converge to the speed at t = 1
which we know is -2 pi
so try the part iii for example
s(1.01 ) = 2 sin(1.01 pi ) + 2 cos(1.01 pi)
= -0.0628 -1.999 = -2.062
s(1) = 2(0)+2(-1) = -2
so
s(1.01)-s(1) = -.062
divide by time of 0.01 seconds
v average = -6.1813
the exact from the second part was -2 pi = - -6.2318 .......
if you did 1 to 1.0001 seconds it would be even closer