Question

The roller coaster car travels down the helical path at constant speed such that the parametric equations that define its position are x = c sin kt, y = c cos kt, z = h − bt, where c, h, and b are constants. Determine the magnitudes of its velocity and acceleration.

Answer 1

Answer:

$\left \| v(t) \right \|=\sqrt{c^2k^2+b^2}$

$\left \| a(t) \right \|=ck^2$

Step-by-step explanation:

Let $r(t)$ denotes position of the roller coaster.

Let $r(t)=\left ( c\sin kt,c\cos kt,h-bt \right )$

Differentiate with respect to t

$v(t)=r'(t)=\left ( ck\cos kt,-ck\sin kt,-b \right )$

Here, $v(t)$ denotes velocity.

Differentiate again with respect to t

$a(t)=v'(t)=\left ( -ck^2\sin kt,-ck^2\cos kt,0 \right )$

Here, $a(t)$ denotes acceleration.

Magnitude of velocity:

$\left \| v(t) \right \|=\sqrt{c^2k^2\cos ^2kt+c^2k^2\sin ^2kt+b^2}=\sqrt{c^2k^2\left ( \cos ^2kt+\sin ^2kt \right )+b^2}=\sqrt{c^2k^2+b^2}$

Magnitude of acceleration:

$\left \| a(t) \right \|=\sqrt{c^2k^4\sin ^2kt+c^2k^4\cos ^2kt+0^2}=\sqrt{c^2k^4\left ( \cos ^2kt+\sin ^2kt \right )}=\sqrt{c^2k^4}=ck^2$