Question

hw 12 - higher order derivatives section 2.7: problem 11 (1 point) the function s(t) describes the position of a particle moving along a coordinate line, where s is in feet and t in seconds. s(t)=t^4 - 72t^2 + 1296, t≥0 if appropriate, enter answers in radical form. use inf to represent ∞. (a) find the velocity and acceleration functions. v(t)= a(t)= (b) find the position, velocity, speed, and acceleration at t = 5. position (ft): velocity (ft/sec): acceleration (ft/sec²):

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the first - derivative of the position function $s(t)$. Using the power rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, if $s(t)=t^{4}-72t^{2}+1296$, then $v(t)=\frac{d}{dt}(t^{4}-72t^{2}+1296)=4t^{3}-144t$.

Step2: Find the acceleration function

The acceleration function $a(t)$ is the first - derivative of the velocity function or the second - derivative of the position function. Differentiating $v(t)=4t^{3}-144t$ using the power rule, we get $a(t)=\frac{d}{dt}(4t^{3}-144t)=12t^{2}-144$.

Step3: Find the position at $t = 5$

Substitute $t = 5$ into $s(t)$: $s(5)=5^{4}-72\times5^{2}+1296=625-72\times25 + 1296=625-1800 + 1296=121$.

Step4: Find the velocity at $t = 5$

Substitute $t = 5$ into $v(t)$: $v(5)=4\times5^{3}-144\times5=4\times125-720=500 - 720=-220$.

Step5: Find the acceleration at $t = 5$

Substitute $t = 5$ into $a(t)$: $a(5)=12\times5^{2}-144=12\times25-144=300 - 144 = 156$.

Answer:

(a) $v(t)=4t^{3}-144t$, $a(t)=12t^{2}-144$
(b) Position (ft): $121$, Velocity (ft/sec): $-220$, Acceleration (ft/sec²): $156$