Question

The relation between time and distance is t = $\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. The retardation is

• A. 2$\alpha v^3$
• B. 2$\beta v^3$
• C. 2$\alpha \beta v^3$
• D. 2${\beta }^2{v}^3$

Answer 1

Answer: (A)

2$\alpha v^3$

The relation between time and the distance is given by

$t=\alpha x^2+\beta x$

Differentiate the above equation with respect to time,

$1=2x\alpha \frac{dx}{dt}+\beta \frac{dx}{dt}$. . . . . . .(1)

Differentiate with respect to time,

$0=2\alpha (x\frac{d{x}^2}{d{t}^2}+\frac{dx}{dt}\frac{dx}{dt})+\beta \frac{d{x}^2}{d{t}^2}$. . . . .(2)

Velocity, $v=\frac{dx}{dt}$. . . . . . .(3)

Acceleration, $a=\frac{d^{2}x}{d{t}^2}$. . . . . . . .(4)

Substitute equation (3) and (4) in equation (1), we get

$0=2\alpha (xa+v^2)+\beta a$

$\beta a=-2\alpha (xa+v^2)=-2\alpha xa-2\alpha v^2$

$\beta a+2\alpha xa=-2\alpha v^2$

$a=-\frac{2\alpha v^2}{\beta +2\alpha x}$. . . . . . . . .(5)

Substitute equation (3) and (4) in equation (1), we get

$1=2x\alpha v+\beta v$

$2x\alpha v=1-\beta v$

$2\alpha x=\frac{1-\beta v}{v}$. . . . . . . . .(6)

Substitute equation (6) in equation (5), we get

$a=-\frac{2\alpha v^2}{\beta +\frac{1-\beta v}{v}}=-\frac{2\alpha v^2}{\beta v+1-\beta v}\times v$

$a=-2\alpha v^3$

The retardation is $2\alpha v^3$.

The correct option is A.