NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Physics
- Chapter
- Motion in Two Dimensions
NTA Abhyas 2020 · Physics · Motion in Two Dimensions
A particle moves in space along the path z = ax^ 3 + by^ 2 in such...
A particle moves in space along the path in such a way that where , and are constants. The acceleration of the particle is
Options
- A\left(6 a c^{2} x + 2 b c^{2}\right) \overset{^}{k}
- B\left(2 a x^{2} + 6 b y^{2}\right) \overset{^}{k}
- C\left(4 b c^{2} x + 3 a c^{2}\right) \overset{^}{k}
- D\left(b c^{2} x + 2 b y\right) \overset{^}{k}
Correct answer
A. \left(6 a c^{2} x + 2 b c^{2}\right) \overset{^}{k}
Step-by-step solution
Given that
∴ \dfrac{{d }^{2} x}{d t^{2}} = \dfrac{{d }^{2} y}{d t^{2}} = 0
Further z = ax3 + by2
∴ \dfrac{d z}{d t} = 3 a x^{2} \dfrac{d x}{d t} + 2 b y \dfrac{d y}{d t}
~ & & & & & = 3 a {c x}^{2} + 2 b c y \left(\dfrac{d x}{d t} =c = \dfrac{d y}{d t}\right)<br /><br />∴ 6 a c x \left(\dfrac{d x}{d t}\right) + 2 b c \left(\dfrac{d y}{d t}\right)\overset{→}{a} = \dfrac{{d }^{2} x}{d t^{2}} \overset{^}{i} + \dfrac{{d }^{2} y}{d t^{2}} \overset{^}{j} + \dfrac{{d }^{2} z}{d t^{2}} \overset{^}{k}<br /><br />~ &= \left(6 a c^{2} x + 2 b c^{2}\right) \overset{^}{k}
∴ \dfrac{{d }^{2} x}{d t^{2}} = \dfrac{{d }^{2} y}{d t^{2}} = 0
Further z = ax3 + by2
∴ \dfrac{d z}{d t} = 3 a x^{2} \dfrac{d x}{d t} + 2 b y \dfrac{d y}{d t}
~ & & & & & = 3 a {c x}^{2} + 2 b c y \left(\dfrac{d x}{d t} =c = \dfrac{d y}{d t}\right)<br /><br />∴ 6 a c x \left(\dfrac{d x}{d t}\right) + 2 b c \left(\dfrac{d y}{d t}\right)\overset{→}{a} = \dfrac{{d }^{2} x}{d t^{2}} \overset{^}{i} + \dfrac{{d }^{2} y}{d t^{2}} \overset{^}{j} + \dfrac{{d }^{2} z}{d t^{2}} \overset{^}{k}<br /><br />~ &= \left(6 a c^{2} x + 2 b c^{2}\right) \overset{^}{k}