(B) The position vector of the particle is given by $\vec{r}(t) = (3t)\hat{i} + (4t^2)\hat{j}$.To find the velocity vector $\vec{v}(t)$,we differentia

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) The position vector of the particle is given by $\vec{r}(t) = (3t)\hat{i} + (4t^2)\hat{j}$.To find the velocity vector $\vec{v}(t)$,we differentiate the position vector with respect to time $t$:$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt} [(3t)\hat{i} + (4t^2)\hat{j}]$$\vec{v}(t) = 3\hat{i} + (8t)\hat{j}$.Now,substitute $t = 2 \ s$ into the velocity equation:$\vec{v}(2) = 3\hat{i} + (8 \times 2)\hat{j}$$\vec{v}(2) = 3\hat{i} + 16\hat{j} \ m/s$.

Class 11 Physics 3-2.Motion in Plane Motion In Two And Three Dimension

Easy

If the position vector of a particle is given by $\vec{r}(t) = (3t)\hat{i} + (4t^2)\hat{j}$,then obtain its velocity vector at $t = 2 \ s$.

A
$3\hat{i} + 8\hat{j}$
B
$3\hat{i} + 16\hat{j}$
C
$6\hat{i} + 16\hat{j}$
D
$3\hat{i} + 32\hat{j}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

3-2.Motion in Plane

Subtopic

Motion In Two And Three Dimension

$A$ hall has the dimensions $10\,m \times 12\,m \times 14\,m$. $A$ fly starting at one corner ends up at a diametrically opposite corner. What is the magnitude of its displacement in meters?

Medium

View Solution

$A$ particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}$,where $a, b,$ and $c$ are constants. The acceleration of the particle is:

Difficult

View Solution

$A$ man moves in the $x-y$ plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector? The man starts from point $P$.

Medium

View Solution

The position of a particle moving in the $xy$-plane at any time $t$ is given by $x = (3t^2 - 6t) \text{ m}$ and $y = (t^2 - 2t) \text{ m}$. Select the correct statement about the moving particle from the following:

Medium

View Solution

$A$ particle is moving in the $xy$-plane such that its position coordinates are given by $x = (4t + t^2) \text{ m}$ and $y = (2t + \frac{t^2}{2}) \text{ m}$,where $t$ is in seconds. What is the velocity of the particle?

EasyTS EAMCET 2022

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If the position vector of a particle is given by $\vec{r}(t) = (3t)\hat{i} + (4t^2)\hat{j}$,then obtain its velocity vector at $t = 2 \ s$.

Similar Questions

$A$ hall has the dimensions $10\,m \times 12\,m \times 14\,m$. $A$ fly starting at one corner ends up at a diametrically opposite corner. What is the magnitude of its displacement in meters?

$A$ particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}$,where $a, b,$ and $c$ are constants. The acceleration of the particle is:

$A$ man moves in the $x-y$ plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector? The man starts from point $P$.

The position of a particle moving in the $xy$-plane at any time $t$ is given by $x = (3t^2 - 6t) \text{ m}$ and $y = (t^2 - 2t) \text{ m}$. Select the correct statement about the moving particle from the following:

$A$ particle is moving in the $xy$-plane such that its position coordinates are given by $x = (4t + t^2) \text{ m}$ and $y = (2t + \frac{t^2}{2}) \text{ m}$,where $t$ is in seconds. What is the velocity of the particle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module