Question

The motion of a particle along a straight line is described by the function $x=6+4t^2–t^4$ where $x$ is in meters and $t$ is in seconds. What is the maximum velocity attained by the particle ?

• A. $\frac{16}{3}\sqrt{\frac{2}{3}}\text{m/s}$
• B. $\frac{14}{3}\sqrt{\frac{2}{3}}\text{m/s}$
• C. $\frac{14}{3}\sqrt{\frac{3}{2}}\text{m/s}$
• D. $\frac{16}{3}\sqrt{\frac{3}{2}}\text{m/s}$

Answer 1

The correct option is A $\frac{16}{3}\sqrt{\frac{2}{3}}\text{m/s}$

We know that velocity is rate of change of position with respect to time, function of position is given as $x=6+4t^2–t^4$

$\therefore$Velocity, $v=\frac{dx}{dt}=8t-4t^3$
acceleration $a=\frac{dv}{dt}=8-12t^2$
For finding the maxima of any function we diffrentiate it and equate to zero.

$\Rightarrow$Velocity is maximum when $\frac{dv}{dt}=0$
$\Rightarrow 8-12t^2=0\Rightarrow t=\sqrt{\frac{2}{3}}$

Now by substituting the time in the function of velocity we get,
maximum velocity,
$=8\sqrt{\frac{2}{3}}-4.\frac{2}{3}.\sqrt{\frac{2}{3}}$
$=\frac{16}{3}\sqrt{\frac{2}{3}}\text{m/s}$

Hence option A is the correct answer