Question

suppose that the equation of motion for a particle (where s is in meters and t in seconds) is (s=(1/3)t^{3}-2t^{2}+4t + 5). (a) find the velocity and acceleration as functions of t. velocity at time (t=t^{2}-4t + 4). acceleration at time (t=). (b) find the acceleration after 1 second. acceleration after 1 second: 1. (c) find the acceleration at the instant when the velocity is 0. acceleration: . video example: solving a similar problem

Explanation:

Step1: Recall the derivative rules

Velocity $v(t)$ is the first - derivative of position $s(t)$. Given $s(t)=\frac{1}{3}t^{3}-2t^{2}+4t + 5$, using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=t^{2}-4t + 4$.

Step2: Find the acceleration function

Acceleration $a(t)$ is the first - derivative of velocity $v(t)$ or the second - derivative of position $s(t)$. Since $v(t)=t^{2}-4t + 4$, then $a(t)=v^\prime(t)=2t-4$.

Step3: Find acceleration at $t = 1$

Substitute $t = 1$ into the acceleration function $a(t)$. So $a(1)=2\times1-4=-2$.

Step4: Find when velocity is 0

Set $v(t)=t^{2}-4t + 4 = 0$. Factor the quadratic equation: $(t - 2)^{2}=0$, so $t = 2$.

Step5: Find acceleration when velocity is 0

Substitute $t = 2$ into the acceleration function $a(t)$. So $a(2)=2\times2-4=0$.

Answer:

(a) Acceleration at time $t=2t - 4$
(b) Acceleration after 1 second: $-2$
(c) Acceleration: $0$