NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Physics
- Chapter
- Motion in One Dimension
NTA Abhyas 2020 · Physics · Motion in One Dimension
A particle starts from the origin of coordinates at time t = 0 and...
A particle starts from the origin of coordinates at time and moves in the plane with a constant acceleration α in the -diretion. Its equation of motion is . Its velocity component in the -direction is
Options
- AVariable
- B
- C
- D
Correct answer
D.
Step-by-step solution
Equation of motion is y = β x^{2}
\dfrac{d^{2} y}{{dt}^{2}} = α
from
\dfrac{\text{dy}}{\text{dt}} = 2 x \dfrac{dx}{\text{dt}} β
\dfrac{d^{2} y}{{dt}^{2}} = \left(2 {\left(\dfrac{\text{dx}}{\text{dt}}\right)}^{2} + \text{2x} \dfrac{d^{2} x}{{dt}^{2}}\right) β
{a}_{y} = \left(2 {\left(\dfrac{\text{dx}}{\text{dt}}\right)}^{2} + 0\right) β
{a}_{y} = 2 {v}_x^{2} β
{v}_{x} = \sqrt{\dfrac{{a}_{y}}{2 β}} = \sqrt{\dfrac{α}{2 β}}
\dfrac{d^{2} y}{{dt}^{2}} = α
from
\dfrac{\text{dy}}{\text{dt}} = 2 x \dfrac{dx}{\text{dt}} β
\dfrac{d^{2} y}{{dt}^{2}} = \left(2 {\left(\dfrac{\text{dx}}{\text{dt}}\right)}^{2} + \text{2x} \dfrac{d^{2} x}{{dt}^{2}}\right) β
{a}_{y} = \left(2 {\left(\dfrac{\text{dx}}{\text{dt}}\right)}^{2} + 0\right) β
{a}_{y} = 2 {v}_x^{2} β
{v}_{x} = \sqrt{\dfrac{{a}_{y}}{2 β}} = \sqrt{\dfrac{α}{2 β}}