Question

If the distance travelled by a particle is $\sqrt{{pt}^2+2qt+r}$ then the acceleration is proportional to

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

Answer 1

The correct option is A $\frac{1}{x}$

$x=\sqrt{{pt}^2+2qt+r}$

We know that

$\frac{dx}{dt}=\frac{1}{2\sqrt{{pt}^2+2qt+r}}\times (2pt+2q)$

$\frac{dx}{dt}=\frac{(pt+q)}{2\sqrt{{pt}^2+2qt+r}}=\frac{pt+q}{2\sqrt{{pt}^2+2qt+r}}$

acceleration $=-\frac{\sqrt{p{t}^2+2q+n}.p-(pt+q)}{(p{t}^2+2q+n{)}^2}.\frac{(pt+q)}{2\sqrt{{pt}^2+2qt+r}}$

$=\frac{(p{t}^2+2q+n)p-(pt+q{)}^2}{(p{t}^2+2q+n)3/2}$

$\frac{d^{2}x}{d{t}^2}=\frac{(p{t}^2+2q+n)p-(pt+q{)}^2}{x^3}$

$\frac{d^{2}x}{d{t}^2}\propto \frac{1}{x^3}$

This is the required proportionality of acceleration.