Question

the position vector of a particle changes with time according to the relation: $r(t)=15t^2 \boldsymbol{i}+(4-20t^2)\boldsymbol{j}$. find the velocity of the particle at any time, t.$15t^2 \boldsymbol{i}+(4-40)\boldsymbol{j}$$30\boldsymbol{i}+(4-20t^2)\boldsymbol{j}$$30t\boldsymbol{i}+(-40t)\boldsymbol{j}$$30t\boldsymbol{i}+(4-40t)\boldsymbol{j}$

Explanation:

Step1: Recall velocity definition

Velocity $\vec{v}(t)$ is derivative of position $\vec{r}(t)$:
$$\vec{v}(t) = \frac{d\vec{r}(t)}{dt}$$

Step2: Differentiate i-component

Differentiate $15t^2$ with respect to $t$:
$$\frac{d}{dt}(15t^2) = 30t$$

Step3: Differentiate j-component

Differentiate $4-20t^2$ with respect to $t$:
$$\frac{d}{dt}(4-20t^2) = -40t$$

Step4: Combine components

Assemble the velocity vector:
$$\vec{v}(t) = 30t\vec{i} -40t\vec{j}$$

Answer:

30ti+(-40t)j