Question

A particle moves along a straight line as per equation $x^2=\alpha t^2+2\beta t+\gamma$, where $x$ is the distance travelled and $\alpha ,\beta ,\gamma$ are constants. Its acceleration varies as

• A. $x^{-3}$
• B. $x^{3/2}$
• C. $x^{-2/3}$
• D. $x^2$

Answer 1

Answer: (A)

$x^{-3}$

$x^2={\alpha t}^2+2\beta t+\gamma 2x\frac{dx}{dt}=2\alpha t+2\beta xv=\alpha t+\beta v\frac{dx}{dt}+x\frac{dv}{dt}=\alpha v^2+xa=\alpha a=\frac{\alpha -v^2}{x}a=\frac{\alpha -{\left\{\right(\alpha t+\beta \left)/x\right\}}^2}{x}i.e.,aisproportionalto{x}^{-3}$