A particle covers 50 m distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed

The correct option is 200 m.

Step-by-step Explanation:

When a particle is projected with an initial velocity on a rough surface, it decelerates and eventually comes to rest due to the force of friction. We can find the relationship between the initial speed and the distance covered before stopping by using the third equation of motion:

$v^2=u^2+2as$

Where:
• $u$ is the initial speed of the particle,
• $v$ is the final speed of the particle (which is 0 since the particle eventually stops),
• $a$ is the constant deceleration (retardation due to friction on the same surface), and
• $s$ is the distance covered before coming to rest.

Substituting $v=0$ into the equation, we get:

$0=u^2-2as$

Rearranging the equation to solve for the stopping distance $s$:

$s=\frac{u^2}{2a}$

Since the surface remains the same, the retardation $a$ remains constant. Therefore, the stopping distance $s$ is directly proportional to the square of the initial speed $u$:

$s\propto u^2$

Let $s_1$ be the initial stopping distance corresponding to initial speed $u_1$, and $s_2$ be the new stopping distance when the speed is doubled ($u_2=2u_1$). We can write the ratio of the two distances as:

$\frac{s_2}{s_1}={\left(\frac{u_2}{u_1}\right)}^2$

Substitute the given values into the ratio equation, where $s_1=50\text{m}$ and $u_2=2u_1$:

$\frac{s_2}{50}={\left(\frac{2u_1}{u_1}\right)}^2$

$\frac{s_2}{50}={\left(2\right)}^2$

$\frac{s_2}{50}=4$

Now, calculate the new distance $s_2$:

$s_2=4\times 50=200\text{m}$

Thus, when projected with double the initial speed, the particle will cover a distance of 200 m before coming to a stop.