Answer:
[tex]\left \| v(t) \right \|=\sqrt{c^2k^2+b^2}[/tex]
[tex]\left \| a(t) \right \|=ck^2[/tex]
Step-by-step explanation:
Let [tex]r(t)[/tex] denotes position of the roller coaster.
Let [tex]r(t)=\left ( c\sin kt,c\cos kt,h-bt \right )[/tex]
Differentiate with respect to t
[tex]v(t)=r'(t)=\left ( ck\cos kt,-ck\sin kt,-b \right )[/tex]
Here, [tex]v(t)[/tex] denotes velocity.
Differentiate again with respect to t
[tex]a(t)=v'(t)=\left ( -ck^2\sin kt,-ck^2\cos kt,0 \right )[/tex]
Here, [tex]a(t)[/tex] denotes acceleration.
Magnitude of velocity:
[tex]\left \| v(t) \right \|=\sqrt{c^2k^2\cos ^2kt+c^2k^2\sin ^2kt+b^2}=\sqrt{c^2k^2\left ( \cos ^2kt+\sin ^2kt \right )+b^2}=\sqrt{c^2k^2+b^2}[/tex]
Magnitude of acceleration:
[tex]\left \| a(t) \right \|=\sqrt{c^2k^4\sin ^2kt+c^2k^4\cos ^2kt+0^2}=\sqrt{c^2k^4\left ( \cos ^2kt+\sin ^2kt \right )}=\sqrt{c^2k^4}=ck^2[/tex]
