JEE Main 202228 JulyEvening Shift
- Exam
- JEE Main
- Year
- 2022
- Date
- 28 July
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Motion in Two Dimensions
JEE Main 2022 · 28 July · Evening Shift · Physics · Motion in Two Dimensions
At time t = 0 a particle starts travelling from a height 7 ^ z cm...
At time a particle starts travelling from a height 7\overset{^}{z}cm in a plane keeping coordinate constant. At any instant of time, it's position along the and directions are defined as and respectively. At acceleration of the particle will be
Options
- A
- B
- C
- D3x + 15y + 7\overset{^}{z}
Correct answer
B.
Step-by-step solution
The position vector of the particle can be written as,
\overset{\to}{r} = 3t\overset{^}{i} + 5{t}^{3}\overset{^}{j} + 7\overset{^}{k}
The velocity of the particle will be,
\overset{\to}{v} = \dfrac{d \overset{\to}{r}}{d t} = 3\overset{^}{i} + 15{t}^{2}\overset{^}{j}
Now the acceleration of the particle will be,
\overset{\to}{a} = \dfrac{d \overset{\to}{v}}{d t} = \dfrac{{d}^{2}\overset{\to}{r}}{d{t}^{2}} = 30t\overset{^}{j}
At t = 1s ⇒ \dfrac{{d}^{2}r}{d{t}^{2}} = 30\overset{^}{j}