Question

a particle starts from rest and moves along the x axis from the origin with its velocity given by v=(18t² - 2.0t³) cm/sec, where t is measured in s. what is the displacement of the particle at the second instant when its acceleration is zero (besides a = 0.0 when t = 0.0)? show all calculations in the answer box.

Explanation:

Step1: Find the acceleration function

Acceleration $a$ is the derivative of velocity $v$ with respect to time $t$. Given $v = 18t^{2}-2.0t^{3}$, then $a=\frac{dv}{dt}=36t - 6t^{2}$ (using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$).

Step2: Find the times when acceleration is zero

Set $a = 0$, so $36t-6t^{2}=0$. Factor out $6t$: $6t(6 - t)=0$. This gives two solutions for $t$: $t = 0$ and $t=6\ s$. We are interested in $t = 6\ s$.

Step3: Find the displacement function

Displacement $x$ is the integral of velocity $v$ with respect to time $t$. So $x=\int vdt=\int(18t^{2}-2.0t^{3})dt$. Using the power - rule for integration $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $x = 18\times\frac{t^{3}}{3}-2\times\frac{t^{4}}{4}+C=6t^{3}-\frac{1}{2}t^{4}+C$. Since the particle starts from the origin ($x = 0$ at $t = 0$), $C = 0$.

Step4: Calculate the displacement at $t = 6\ s$

Substitute $t = 6\ s$ into the displacement function $x=6t^{3}-\frac{1}{2}t^{4}$. Then $x=6\times6^{3}-\frac{1}{2}\times6^{4}=6^{4}-\frac{1}{2}\times6^{4}=\frac{1}{2}\times6^{4}=648\ cm$.

Answer:

$648\ cm$