Question

The position of an object is given by r=(9ti^+4t3j^)m /s find its velocity at time t=1

Answer 1

Given conditions ⇒

Velocity of the object = $9t\hat{i}+ 12t\hat{j}}$

∴ Magnitude of the Position = $\sqrt{(9t)^2 + (12t)^2}$

= $\sqrt{81t^2 + 144t^2}$

= $\sqrt{t^2(81 + 144)}$

= $\sqrt{t^2(225)}$

= 15t

Now, the Velocity of the object is 15(1) = 15 m/s.

Hope it helps.

Answer 2

The position of an object is given by
$\bf{\vec{r}=(9t\hat{i}+4t^3\hat{j})m/s}$

we can separate in x and y - components.
e.g.,$\bf{r_x=9t}$
differentiate it with respect to t,
$\bf{\frac{dr_x}{dt}=v_x=9}$

similarly, $\bf{r_y=4t^3}$
differentiate with respect to t,
$\bf{\frac{dr_y}{dt}=v_y=4.3t^{3-1}=12t^2}\\\\\bf{v_y(at\:t=1)=12(1)^2=12}$

now, velocity , $\bf{\vec{v}=9\hat{i}+12\hat{j}}$
and magnitude of v =$\bf{\sqrt{9^2+12^2}}=15m/s$