Question

The displacement of a particle moving in a straight line is described by the relation $x\text{(in m)}=6+12t-2t^2$. The distance covered by the particle in first $5\text{s}$ is

• A. $16\text{m}$
• B. $10\text{m}$
• C. $24\text{m}$
• D. $26\text{m}$

Answer 1

The correct option is D $26\text{m}$

In order to calculate distance, we need to calculate displacement from the position time function of the particle. When a particle moves without changing its direction of motion, distance travelled is equal to the magnitude of displacement.
Using this we will calculate if particle has changed its direction in the time interval $0$ to $5\text{s}$ and correspondingly calculate the distance from the displacement.
In order to check for direction of motion, velocity is considered.
$x\left(t\right)=6+12t-2t^2$
$v=\frac{dx}{dt}=12-4t=0$
$t=3\text{s}$
Velocity is positive before $t=3\text{s}$ and negative after that.
Distance $(t=0$ to $t=5\text{s})$
$=|\text{Displacement}(t=0\text{to}t=3s)|+|\text{Displacement}(t=3\text{s to}t=5\text{s})|$
$=|(x_3-x_0)|+|x_5-x_3|$
$=|(6+36-18)-\left(6\right)|+|(6+60-50)-(6+36-18)|$
$=|24-6|+|16-24|=26\text{m}$