A particle moves in 2D motion such that its position and time related to each other as - - 4t2 - 4t-3 - - - 2- - 1- Find velocity of particle at - - 1 sec in m-s

Given: x=4t^2+4t+3⇒ v_x=dxdt=8t+4 so, v_x|_@ t=1 sec=8×1+4=12 Also, y=2t 1⇒ v_y=dydt=2 So, at t=1 sec, v_y=2 Thus, velocity of particle is given by v=v_xî+v_yĵ ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Instantaneous Velocity

A particle mo...

Question

A particle moves in $2 D$ motion such that its position and time related to each other as $x = 4 t^{2} + 4 t + 3, y = 2 t - 1$ . Find velocity of particle at $t = 1 sec (in m/s)$

\sqrt{158}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{140}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{148}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{144}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{148}$
Given:
$x = 4 t^{2} + 4 t + 3 \Rightarrow v_{x} = \frac{d x}{d t} = 8 t + 4$
so, $v_{x} |_{@ t = 1 sec} = 8 \times 1 + 4 = 12$
Also, $y = 2 t - 1 \Rightarrow v_{y} = \frac{d y}{d t} = 2$
So, at $t = 1 sec$ , $v_{y} = 2$
Thus, velocity of particle is given by
$\to v = v_{x}^i + v_{y}^j$
$| \to v | = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{12^{2} + 2^{2}} = \sqrt{148}$
Hence, option $(B)$ is correct.

Suggest Corrections

Similar questions

Q. A particle moves in

2 D

motion such that its position and time related to each other as

x = t^{2} + 10, y = t - 1

. Find velocity of particle at

t = 2 sec (in m/s)

Q. A particle starts its

2 D

motion from origin such that its velocity in

x -

direction remains constant at

3 m/s

and in

y -

direction its acceleration varies with time as

a_{y} = (t^{2} + 2) {m/s}^{2}

. Find the position vector of particle at

t = 2 sec

Q. The

x

and

y

coordinates of the particle at any time are

x = 2 t^{2}

and

y = 10 t

respectively, where

x

and

y

are in meters and

t

in seconds. The acceleration of the particle at

t = 2 sec

({in m/s}^{2})

Q. a particle is moving in a straight line such that its displacements is related to time as X=3-4t-5t^2 m . the correct velocity time v-t graph will be

Q. A particle executes the motion described by

x (t) = x_{0} (1 - e^{- γ t}); t \geq 0, x_{0} < 0.

Find maximum and minimum values of 𝑥(𝑡), 𝑣(𝑡), 𝑎(𝑡). Show that 𝑥(𝑡) and 𝑎(𝑡) increase with time and 𝑣(𝑡) decreases with time.

Join BYJU'S Learning Program

A particle moves in 2D motion such that its position and time related to each other as x=4t2+4t+3,y=2t−1. Find velocity of particle at t=1 sec (in m/s)

The correct option is C √148Given: x=4t2+4t+3⇒ vx=dxdt=8t+4 so, vx|@ t=1 sec=8×1+4=12 Also, y=2t−1⇒ vy=dydt=2 So, at t=1 sec, vy=2 Thus, velocity of particle is given by →v=vx^i+vy^j |→v|=√v2x+v2y=√122+22=√148 Hence, option (B) is correct.

A particle moves in $2 D$ motion such that its position and time related to each other as $x = 4 t^{2} + 4 t + 3, y = 2 t - 1$ . Find velocity of particle at $t = 1 sec (in m/s)$