Question

For a particle displacement x at time t is given by $x^2=a{t}^2+b$ where a and b are constants. Its acceleration at time t is proportional to

• A. $\frac{1}{x^3}$
• B. $\frac{1}{x}-\frac{1}{x^2}$
• C. $-\frac{t}{x^2}$
• D. $\frac{1}{x}-\frac{t^2}{x^3}$

Answer 1

The correct option is A

$\frac{1}{x^3}$

$x^2=a{t}^2+b$

$2x\frac{dx}{dt}=2at$

$xv=at(\because v=\frac{dx}{dt})$

$x\frac{dv}{dt}+v\frac{dx}{dt}=aorx\frac{dv}{dt}=a-v^2$

$\frac{dv}{dt}=\frac{a-v^2}{x}=\frac{a-{\left(\frac{at}{x}\right)}^2}{x}=\frac{a-\frac{a^2{t}^2}{x^2}}{x}=\frac{a(x^2-a{t}^2)}{x^3}$

$\frac{dv}{dt}=\frac{ab}{x^3}$

Acceleration $\propto \frac{1}{x^3}$