Trajectory of a particle moving in vast maidan is as shown in figure. coordinates of a position A ar

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3-2.Motion in Plane

medium

The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are $(0,2)$. The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are

$(1, 4)$

$(5, 3)$

$(3, 4)$

$(4, 1)$

Solution

The coordinates of position $A$ ar $(0,2)$ is same as the average velocity between the points are $(5,3)$ in graph of a moving particle as shown in fig.Because both have same slope.

Standard 11

Physics

A person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?

normal

(KVPY-2016)

The coordinates of a moving particle at any time $t$ are given by $x = a\, t^2$ and $y = b\, t^2$. The speed of the particle is

easy

(AIIMS-2012)

The position vector of a particle changes with time according to the relation $\vec r\left( t \right) = 15{t^2}\hat i + \left( {4 – 20{t^2}} \right)\hat j$. What is the magnitude of the acceleration at $t = 1$ ?

medium

(JEE MAIN-2019)

The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) m$ at $t=$ $2 s$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take $g =10\,ms ^{-2}$ )

medium

A particle moves in $x-y$ plane with velocity $\vec v = a\widehat i\, + \,bx\widehat j$ where $a$ & $b$ are constants. Initially particle was at origin then trajectory equation is:-

hard

The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are $(0,2)$. The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are

Solution

Similar Questions

A person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?

The coordinates of a moving particle at any time $t$ are given by $x = a\, t^2$ and $y = b\, t^2$. The speed of the particle is

The position vector of a particle changes with time according to the relation $\vec r\left( t \right) = 15{t^2}\hat i + \left( {4 – 20{t^2}} \right)\hat j$. What is the magnitude of the acceleration at $t = 1$ ?

The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) m$ at $t=$ $2 s$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take $g =10\,ms ^{-2}$ )

A particle moves in $x-y$ plane with velocity $\vec v = a\widehat i\, + \,bx\widehat j$ where $a$ & $b$ are constants. Initially particle was at origin then trajectory equation is:-

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