Which physical quantity can be found by first differntiation and second differentiation of position vector ?
Which physical quantity can be found by first differntiation and second differentiation of position vector ?
Similar Questions
A car travels $6\, km$ towards north at an angle of $45^o$ to the east and then travels distance of $4\, km$ towards north at an angle $135^o$ to east. How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east?
A car travels $6\, km$ towards north at an angle of $45^o$ to the east and then travels distance of $4\, km$ towards north at an angle $135^o$ to east. How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east?
- [AIIMS 2008]
Give examples of one dimensional, two dimensional and three dimensional motion.
Give examples of one dimensional, two dimensional and three dimensional motion.
Starting from the origin at time $t=0,$ with initial velocity $5 \hat{ j }\, ms ^{-1},$ a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{ i }+4 \hat{ j })\, ms ^{-2}$. At time $t$, its coordinates are $\left(20\, m , y _{0}\, m \right) .$ The values of $t$ and $y _{0},$ are respectively
Starting from the origin at time $t=0,$ with initial velocity $5 \hat{ j }\, ms ^{-1},$ a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{ i }+4 \hat{ j })\, ms ^{-2}$. At time $t$, its coordinates are $\left(20\, m , y _{0}\, m \right) .$ The values of $t$ and $y _{0},$ are respectively
- [JEE MAIN 2020]
A point moves in $x-y$ plane as per $x=kt,$ $y = kt\left( {1 - \alpha t} \right)$ where $k\,\& \,\alpha \,$ are $+ve$ constants. The equation of trajectory is
A point moves in $x-y$ plane as per $x=kt,$ $y = kt\left( {1 - \alpha t} \right)$ where $k\,\& \,\alpha \,$ are $+ve$ constants. The equation of trajectory is
A particle moves in the $xy$ plane with a constant acceleration $'g'$ in the negative $y$-direction. Its equation of motion is $y = ax-bx^2$, where $a$ and $b$ are constants. Which of the following are correct?
A particle moves in the $xy$ plane with a constant acceleration $'g'$ in the negative $y$-direction. Its equation of motion is $y = ax-bx^2$, where $a$ and $b$ are constants. Which of the following are correct?