$A$ particle moves towards east with velocity $5\, m/s$. After $10\, s$,its direction changes towards north with the same velocity. The average acceleration of the particle is

  • A
    Zero
  • B
    $\frac{1}{\sqrt{2}}\,m/s^2$ in North-West direction
  • C
    $\frac{1}{\sqrt{2}}\,m/s^2$ in North-East direction
  • D
    $\frac{1}{\sqrt{2}}\,m/s^2$ in South-West direction

Explore More

Similar Questions

Motion in two dimensions in a plane can be studied by expressing position,velocity,and acceleration as vectors in Cartesian coordinates $\vec{A} = A_{x} \hat{i} + A_{y} \hat{j}$,where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ and $y$ directions,respectively,and $A_{x}$ and $A_{y}$ are corresponding components of $\vec{A}$. Motion can also be studied by expressing vectors in circular polar coordinates as $\vec{A} = A_{r} \hat{r} + A_{\theta} \hat{\theta}$,where $\hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$ and $\hat{\theta} = -\sin \theta \hat{i} + \cos \theta \hat{j}$ are unit vectors along the directions in which $r$ and $\theta$ are increasing.
$(a)$ Express $\hat{i}$ and $\hat{j}$ in terms of $\hat{r}$ and $\hat{\theta}$.
$(b)$ Show that both $\hat{r}$ and $\hat{\theta}$ are unit vectors and are perpendicular to each other.
$(c)$ Show that $\frac{d}{dt}(\hat{r}) = \omega \hat{\theta}$,where $\omega = \frac{d\theta}{dt}$ and $\frac{d}{dt}(\hat{\theta}) = -\omega \hat{r}$.
$(d)$ For a particle moving along a spiral given by $\vec{r} = a\theta \hat{r}$,where $a = 1$ (unit),find the dimensions of $a$.
$(e)$ Find velocity and acceleration in polar vector representation for a particle moving along the spiral described in $(d)$ above.

Difficult
View Solution

At a given instant of time,two particles have position vectors $4 \hat{i} + 4 \hat{j} + 57 \hat{k} \ m$ and $2 \hat{i} + 2 \hat{j} + 5 \hat{k} \ m$ respectively. If the velocity of the first particle is $0.4 \hat{i} \ ms^{-1}$,what is the velocity of the second particle in $ms^{-1}$ if they collide after $10 \ s$?

At $t = 0$,a projectile is fired from a point $O$ (taken as origin) on the ground with a speed of $50 \ m/s$ at an angle of $53^{\circ}$ with the horizontal. It passes through two points $A$ and $B$,each at a height of $75 \ m$ above the horizontal,as shown in the figure. The horizontal separation between the points $A$ and $B$ is . . . . . . $m$.

In the projectile motion of an object,the object reaches its maximum height where its speed is half of its initial speed. Then the ratio between the range and the maximum height of the projectile is

$A$ ball is projected from the ground at an angle of $\theta$ from the horizontal. The graph of kinetic energy $(K)$ versus time $(t)$ will be:

Difficult
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