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 $x=40+4t-t^4$. Displacement of the particle before coming to rest is

• A. $3\text{m}$
• B. $2\text{m}$
• C. $4\text{m}$
• D. $6\text{m}$

Answer 1

The correct option is A $3\text{m}$

$x=40+4t-t^4$
At $t=0,x_0=40\text{m}$
We know that velocity, $v=\frac{dx}{dt}=4-4t^3$
When particle comes to rest,
$\frac{dx}{dt}=v=0$
$\therefore 4-4t^3=0\Rightarrow t^3=1\Rightarrow t=1\text{seconds}$
At $t=1$ seconds, $x_2=40+4\left(1\right)-1^4=43\text{m}$
$\therefore$ Displacement of the particle before coming to rest is given by $x_2-x_0=43\text{m}-40\text{m}=3\text{m}$
Hence, the correct answer is option (a).