Question

the equation below gives the position s = f(t) of a body moving on a coordinate line (s in meters, t in seconds). find the bodys velocity, speed, acceleration, and jerk at time t = $\frac{pi}{4}$ sec. s = 5 - 3 cos t. the velocity of the body at t = $\frac{pi}{4}$ sec is $\frac{3sqrt{2}}{2}$ m/sec. the speed of the body at t = $\frac{pi}{4}$ sec is m/sec.

Explanation:

Step1: Recall velocity - position relationship

Velocity $v(t)$ is the derivative of position $s(t)$. Given $s = 5-3\cos t$, then $v(t)=\frac{ds}{dt}=3\sin t$.

Step2: Calculate velocity at $t = \frac{\pi}{4}$

Substitute $t=\frac{\pi}{4}$ into $v(t)$. $v(\frac{\pi}{4}) = 3\sin(\frac{\pi}{4})=3\times\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$ m/s.

Step3: Recall speed - velocity relationship

Speed is the absolute - value of velocity. Since $v(\frac{\pi}{4})=\frac{3\sqrt{2}}{2}>0$, the speed at $t = \frac{\pi}{4}$ is the same as the velocity magnitude.

Step4: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Since $v(t)=3\sin t$, then $a(t)=\frac{dv}{dt}=3\cos t$.

Step5: Calculate acceleration at $t=\frac{\pi}{4}$

Substitute $t = \frac{\pi}{4}$ into $a(t)$. $a(\frac{\pi}{4})=3\cos(\frac{\pi}{4})=3\times\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$ m/s².

Step6: Recall jerk - acceleration relationship

Jerk $j(t)$ is the derivative of acceleration $a(t)$. Since $a(t)=3\cos t$, then $j(t)=\frac{da}{dt}=- 3\sin t$.

Step7: Calculate jerk at $t=\frac{\pi}{4}$

Substitute $t=\frac{\pi}{4}$ into $j(t)$. $j(\frac{\pi}{4})=-3\sin(\frac{\pi}{4})=-\frac{3\sqrt{2}}{2}$ m/s³.

Answer:

Speed: $\frac{3\sqrt{2}}{2}$ m/s
Acceleration: $\frac{3\sqrt{2}}{2}$ m/s²
Jerk: $-\frac{3\sqrt{2}}{2}$ m/s³