Question

1. a particle moves along a line with position $x(t)=3t^{2}+2t$ m. find its velocity and acceleration at $t = 2s$.
2. given $x(t)=6t^{2}+7t$ m. find its velocity and acceleration at $t = 5s$.

Explanation:

Step1: Recall velocity - position relationship

Velocity $v(t)$ is the derivative of position $x(t)$. For $x(t)=3t^{2}+2t$, using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(3t^{2}+2t)=6t + 2$.

Step2: Calculate velocity at $t = 2s$

Substitute $t = 2$ into $v(t)$: $v(2)=6\times2+2=12 + 2=14m/s$.

Step3: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Since $v(t)=6t + 2$, then $a(t)=\frac{d}{dt}(6t + 2)=6m/s^{2}$ (constant).

Step4: For the second part, find velocity of $x(t)=6t^{2}+7t$

Using the power - rule, $v(t)=\frac{d}{dt}(6t^{2}+7t)=12t+7$.

Step5: Calculate velocity at $t = 5s$

Substitute $t = 5$ into $v(t)$: $v(5)=12\times5+7=60 + 7=67m/s$.

Step6: Find acceleration of $x(t)=6t^{2}+7t$

Since $v(t)=12t + 7$, then $a(t)=\frac{d}{dt}(12t + 7)=12m/s^{2}$ (constant).

Answer:

1. Velocity at $t = 2s$: $14m/s$, Acceleration at $t = 2s$: $6m/s^{2}$
2. Velocity at $t = 5s$: $67m/s$, Acceleration at $t = 5s$: $12m/s^{2}$