Question

Displacement of a particle in x- y plane varies with time t is given by (t^2i+2j)m. Velocity of the particle at t=1 s is

Answer 1

Answer:

$\boxed{\mathfrak{Velocity \ of \ the \ particle \ (\overrightarrow{v}) = 2 \hat{i} \ m/s}}$

Explanation:

Displacement of particle in x-y plane w.r.t. time is given as:

$\rm \overrightarrow{x} =( {t}^{2} \hat{i} + 2 \hat{j}) \: m$

Rate of change of displacement w.r.t. time is equal to velocity i.e.

$\boxed{ \bold{ \overrightarrow{v} = \frac{d\overrightarrow{x}}{dt} }}$

So,

$\rm \implies \overrightarrow{v} = \dfrac{d\overrightarrow{x}}{dt} \\ \\ \rm \implies \overrightarrow{v} = \dfrac{d}{dt} ( {t}^{2} \hat{i} + 2 \hat{j}) \\ \\ \rm \implies \overrightarrow{v} = 2t \hat{i} \: m {s}^{ - 1}$

At t = 1 s:

$\rm \implies \overrightarrow{v} = (2 \times 1) \hat{i} \: m {s}^{ - 1} \\ \\ \rm \implies \overrightarrow{v} = 2\hat{i} \: m {s}^{ - 1}$

Answer 2

Answer:

x = t²i + 2j

v = dx/dt

v = 2ti

v = 2i (t = 1)