a particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet…

a particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t)=0.01t^4 - 0.02t^3. (a) find the velocity at time t (in ft/s). v(t)= (b) what is the velocity after 1 second(s)? v(1)= ft/s (c) when is the particle at rest? smaller value t = s larger value t = s (d) when is the particle moving in the positive direction? (enter your answer using interval notation.) (e) find the total distance traveled during the first 10 seconds. (round your answer to two decimal places.) (f) find the acceleration at time t (in ft/s^2). a(t)= find the acceleration after 1 second(s). a(1)= ft/s^2 (g) graph the position, velocity, and acceleration functions for the first 10 seconds.

a particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t)=0.01t^4 - 0.02t^3. (a) find the velocity at time t (in ft/s). v(t)= (b) what is the velocity after 1 second(s)? v(1)= ft/s (c) when is the particle at rest? smaller value t = s larger value t = s (d) when is the particle moving in the positive direction? (enter your answer using interval notation.) (e) find the total distance traveled during the first 10 seconds. (round your answer to two decimal places.) (f) find the acceleration at time t (in ft/s^2). a(t)= find the acceleration after 1 second(s). a(1)= ft/s^2 (g) graph the position, velocity, and acceleration functions for the first 10 seconds.

Answer

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v(t)$ is the derivative of displacement $s(t)$. Given $s(t)=0.01t^{4}-0.02t^{3}$, using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(0.01t^{4}-0.02t^{3})$. $v(t)=0.01\times4t^{3}-0.02\times3t^{2}=0.04t^{3}-0.06t^{2}$

Step2: Find velocity at $t = 1$

Substitute $t = 1$ into $v(t)$. $v(1)=0.04\times1^{3}-0.06\times1^{2}=0.04 - 0.06=- 0.02$

Step3: Find when particle is at rest

The particle is at rest when $v(t)=0$. So we set $0.04t^{3}-0.06t^{2}=0$. Factor out $t^{2}$: $t^{2}(0.04t - 0.06)=0$. This gives two solutions: $t^{2}=0\Rightarrow t = 0$ and $0.04t-0.06 = 0\Rightarrow t=\frac{0.06}{0.04}=1.5$

Step4: Find when particle moves in positive direction

The particle moves in the positive direction when $v(t)>0$. Since $v(t)=t^{2}(0.04t - 0.06)$ and $t^{2}\geq0$ for all real $t$, we consider $0.04t - 0.06>0$ (excluding $t = 0$). Solving $0.04t-0.06>0$ gives $t > 1.5$. In interval notation, $(1.5,\infty)$

Step5: Find total distance traveled in first 10 seconds

The total distance $D$ is given by $D=\int_{0}^{10}|v(t)|dt$. Since $v(t)=0.04t^{3}-0.06t^{2}=t^{2}(0.04t - 0.06)$ is negative on $[0,1.5]$ and positive on $[1.5,10]$. $D=-\int_{0}^{1.5}(0.04t^{3}-0.06t^{2})dt+\int_{1.5}^{10}(0.04t^{3}-0.06t^{2})dt$ First integral: $-\left[\frac{0.04}{4}t^{4}-\frac{0.06}{3}t^{3}\right]{0}^{1.5}=-(0.01\times1.5^{4}-0.02\times1.5^{3})$ Second integral: $\left[\frac{0.04}{4}t^{4}-\frac{0.06}{3}t^{3}\right]{1.5}^{10}=(0.01\times10^{4}-0.02\times10^{3})-(0.01\times1.5^{4}-0.02\times1.5^{3})$ $D=(0.01\times10^{4}-0.02\times10^{3})-2(0.01\times1.5^{4}-0.02\times1.5^{3})\approx93.81$

Step6: Recall acceleration - velocity relation

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Since $v(t)=0.04t^{3}-0.06t^{2}$, using the power - rule, $a(t)=\frac{d}{dt}(0.04t^{3}-0.06t^{2})=0.04\times3t^{2}-0.06\times2t=0.12t^{2}-0.12t$

Step7: Find acceleration at $t = 1$

Substitute $t = 1$ into $a(t)$. $a(1)=0.12\times1^{2}-0.12\times1=0$

Answer:

(a) $v(t)=0.04t^{3}-0.06t^{2}$ (b) $v(1)=-0.02$ (c) Smaller value: $t = 0$, Larger value: $t = 1.5$ (d) $(1.5,\infty)$ (e) $93.81$ (f) $a(t)=0.12t^{2}-0.12t$, $a(1)=0$ (g) Graphing is not possible to show in this text - based format. You can use graphing software like Desmos or a graphing calculator to graph $s(t)=0.01t^{4}-0.02t^{3}$, $v(t)=0.04t^{3}-0.06t^{2}$ and $a(t)=0.12t^{2}-0.12t$ for $t\in[0,10]$.