The displacement of a particle is given by x = (t – 2)² where x is in metres and t in seconds. The distance covered by the particle in first 4 seconds is

Correct Answer: 8 m

To find the total distance covered by the particle in the first 4 seconds, we must analyze the motion of the particle and check if it changes direction (i.e., when its velocity becomes zero) during this time interval.

The position (displacement from origin) of the particle as a function of time is given by:
$x={(t-2)}^2$

First, let's find the velocity of the particle by differentiating the position with respect to time:
$v=\frac{dx}{dt}=\frac{d}{dt}{(t-2)}^2=2(t-2)$

The particle momentarily comes to rest and reverses its direction when the velocity is zero:
$v=0\Rightarrow 2(t-2)=0\Rightarrow t=2\text{seconds}$

Since the direction of motion changes at $t=2\text{s}$, which lies within the interval of the first 4 seconds (from $t=0\text{s}$ to $t=4\text{s}$), we must calculate the distance covered in two separate time intervals: from $t=0\text{s}$ to $t=2\text{s}$, and from $t=2\text{s}$ to $t=4\text{s}$.

Let's calculate the position of the particle at $t=0\text{s}$, $t=2\text{s}$, and $t=4\text{s}$:
At $t=0\text{s}$:
$x_0={(0-2)}^2=4\text{m}$
At $t=2\text{s}$:
$x_2={(2-2)}^2=0\text{m}$
At $t=4\text{s}$:
$x_4={(4-2)}^2=4\text{m}$

Now, we find the distance covered in each interval:
Distance from $t=0\text{s}$ to $t=2\text{s}$:
$d_1=|x_2-x_0|=|0-4|=4\text{m}$
Distance from $t=2\text{s}$ to $t=4\text{s}$:
$d_2=|x_4-x_2|=|4-0|=4\text{m}$

The total distance covered by the particle in the first 4 seconds is the sum of these two distances:
$\text{Total Distance}=d_1+d_2=4\text{m}+4\text{m}=8\text{m}$