Question

The displacement of a particle moving in a straight line is given by relation $x=6+12t-2t^2$. Where 't' is in seconds and 'x' is in meters. The distance covered by particle in first $3s$ is

• A. $20m$
• B. $32m$
• C. $24m$
• D. $18m$

Answer 1

The correct option is D $18m$

Given:

The displacement of particle is given by $x=6+12t-2t^2$

To find:

Distance covered by particle in first $3$ sec.

Since the expression is for displacement. So it is necessary to find that the particle does not change its direction.

Time at which the velocity of the particle becomes zero $\to$

$v=\frac{dx}{dt}$

$=\frac{d}{dt}(6+12t-2t^2)$

$=12-4t$

for velocity will be zero $\to 0=12-4t$

$t=3$ sec

So, its direction will chagne after $3$ sec.

So, distance in $3$ sec $\to$

$x_3=6+12t-2t^2$

Put $t=3$ s

$x_3=6+12\left(3\right)-2(3{)}^2$

$=42-18$ m

$=24$ m

position of particle at $t=0$

$x_0=6+12\left(0\right)-2(0{)}^2$

$=6$ m

So, distance covered in $3$ sec

$x=x_3-x_0$

$=24-6$ m

$=18$ m