Question

A particle moves, along a straight line OX. At a time $t$ (in second), the position $x$ (in metre) of the particle from $O$ is given by $x=40+12t-t^3$. How much distance particle covers before coming to rest?

• A. $16\text{m}$
• B. $24\text{m}$
• C. $40\text{m}$
• D. $56\text{m}$

Answer 1

The correct option is A $16\text{m}$

$x=40+12t-t^3$
$\therefore \text{Velocity},v=\frac{dx}{dt}=12-3t^2$
When particle comes to rest,
$\frac{dx}{dt}=v=0$
$\therefore 12-3t^2=03t^2=12\Rightarrow t=2\text{seconds}$
Distance travelled by the particle before coming to rest is
$|x_2-x_0|=|(40+12\times 2-2^3)-(40+12\times 0-0^3)|=16\text{m}$
Hence, the correct answer is option (a).