Answer:
(i) [1,2]
[tex]V_a_v_g=\frac{s(2)-s(1)}{2-1} =\frac{2-(-2)}{1} =4cm/s[/tex]
(ii) [1,1,1]
[tex]V_a_v_g=\frac{s(1.1)-s(1)}{1.1-1} =\frac{-3.447-(-2)}{0.1} =-14.47cm/s[/tex]
(iii)
[tex]V_a_v_g=\frac{s(1.01)-s(1)}{1.01-1} =\frac{-2.156-(-2)}{0.01} =-15.6cm/s[/tex]
(iv)
[tex]V_a_v_g=\frac{s(1.001)-s(1)}{1.001-1} =\frac{-2.016-(-2)}{0.001} =-15.698cm/s[/tex]
(b)
[tex]\frac{ds}{dt} \left \{ {t=1}} \right =-15.7076327\approx-15.7cm/s[/tex]
Step-by-step explanation:
[tex]s(t)=5sin(\pi t)+2cos(\pi t)[/tex]
First. Let's find the displacement for every given time:
t=1s
[tex]s(1)=5sin(\pi)+2cos(\pi)=-2cm=-0.02m[/tex]
t=2s
[tex]s(2)=5sin(\pi *2)+2cos(\pi *2)=2cm=0.02m[/tex]
t=1.1s
[tex]s(1.1)=5sin(\pi *1.1)+2cos(\pi *1.1)\approx -3.447cm\approx-0.03447m[/tex]
t=1.01s
[tex]s(1.101)=5sin(\pi *1.101)+2cos(\pi *1.101)\approx -2.156cm\approx-0.02156m[/tex]
t=1.001s
[tex]s(1.001)=5sin(\pi *1.001)+2cos(\pi *1.001)\approx -2.016cm\approx-0.02016m[/tex]
Now, the average velocity can be found as:
[tex]V_a_v_g=\frac{S_f-S_i}{t_f-t_i}[/tex]
Where:
[tex]S_f=Final\hspace{3}displacement\\S_i=Initial\hspace{3}displacement\\t_f=Final\hspace{3}time\\t_i=Initial\hspace{3}time\\[/tex]
Hence:
(i) [1,2]
[tex]V_a_v_g=\frac{s(2)-s(1)}{2-1} =\frac{2-(-2)}{1} =4cm/s[/tex]
(ii) [1,1,1]
[tex]V_a_v_g=\frac{s(1.1)-s(1)}{1.1-1} =\frac{-3.447-(-2)}{0.1} =-14.47cm/s[/tex]
(iii)
[tex]V_a_v_g=\frac{s(1.01)-s(1)}{1.01-1} =\frac{-2.156-(-2)}{0.01} =-15.6cm/s[/tex]
(iv)
[tex]V_a_v_g=\frac{s(1.001)-s(1)}{1.001-1} =\frac{-2.016-(-2)}{0.001} =-15.698cm/s[/tex]
(b) In order to estimate the instantaneous velocity of the particle for t=1, we need to find its derivative:
[tex]\frac{ds}{dt} \left \{ {t=1}} \right. =5\pi cos(\pi t) -2\pi sin(\pi t)=5\pi cos(\pi (1)) -2\pi sin(\pi (1)) \\\\\frac{ds}{dt} \left \{ {t=1}} \right =-15.7076327\approx-15.7cm/s[/tex]
