The displacement x of a particle moving in one dimension, under the action of a constant force is re

Q: Why did we square both sides of $\sqrt{x} = t - 3$? Doesn't that introduce extraneous roots?

Great question! Squaring can sometimes introduce extraneous solutions, but here $t \ge 3$ is implied by the original equation since $\sqrt{x}$ must be non-negative. However, in kinematics, we often assume the parabolic path $x = (t-3)^2$ holds for all $t \ge 0$ under the action of a constant force. The math perfectly models a particle slowing down, stopping, and reversing direction.

Q: Why is the initial velocity negative?

A negative velocity simply means the particle is moving in the negative $x$-direction (towards the origin). It starts at $x = 9$ m and moves towards $x = 0$ m, so its position is decreasing, hence the negative velocity.

Q: Can I find the work done using $W = F \cdot d$ instead of the Work-Energy Theorem?

Absolutely! Since $v = 2t - 6$, the acceleration is $a = \frac{dv}{dt} = 2 \text{ m/s}^2$. The force is $F = ma = 2m$. The displacement from $t=0$ to $t=6$ is $\Delta x = x(6) - x(0) = 9 - 9 = 0$. Therefore, $W = F \cdot \Delta x = 2m \cdot 0 = 0$. Both methods give the exact same result!

Q: Why is the work done zero if the particle was moving the whole time?

Work done by a force measures the *change* in kinetic energy. Even though the particle traveled a total distance of $18$ m ($9$ m forward, $9$ m backward), its speed at $t=0$ and $t=6$ was exactly the same ($6$ m/s). The force did negative work to slow it down, and then an equal amount of positive work to speed it back up, netting to zero.

Q: What does the graph of this motion look like?

The position-time graph is a parabola opening upwards with its vertex at $(3, 0)$. The velocity-time graph is a straight line with a positive slope, crossing the time axis at $t = 3$ s. This perfectly visualizes constant acceleration!

Question

The displacement $x$ of a particle moving in one dimension, under the action of a constant force is related to the time $t$ by the equation $t = \sqrt{x} + 3$ where $x$ is in metre and $t$ in second.
Find
(a) the displacement of the particle when its velocity is zero, and
(b) the work done by the force in the first 6 s.

Sigma Percentile · Accepted Answer

The problem presents us with a rather peculiar equation:

$$t = \sqrt{x} + 3$$

At first glance, this might look intimidating because time $t$ is given as a function of position $x$. In standard kinematics, we are accustomed to seeing position as a function of time.

### Unveiling the Kinematics

Our first instinct should be to rearrange this equation to isolate $x$. By moving the constant to the other side, we get:

$$\sqrt{x} = t - 3$$

To eliminate the square root, we square both sides of the equation. This yields a beautiful parabolic relationship:

$$x = (t - 3)^2$$

Expanding this expression using the standard algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$, we arrive at the explicit position-time function:

$$x = t^2 - 6t + 9$$

This equation perfectly matches the standard kinematic equation for constant acceleration, $x = ut + \frac{1}{2}at^2 + x_0$. By comparing the coefficients, we can immediately see that the initial position is $x_0 = 9 	ext{ m}$, the initial velocity is $u = -6 	ext{ m/s}$, and the constant acceleration is $a = 2 	ext{ m/s}^2$.

To find the velocity function, we simply take the derivative of the position with respect to time:

$$v = \frac{dx}{dt} = \frac{d}{dt}(t^2 - 6t + 9) = 2t - 6$$

### The Moment of Rest

Part (a) of the question asks for the displacement of the particle when its velocity is zero.

First, we must determine **when** this happens. We set our velocity function to zero:

$$2t - 6 = 0$$

Solving for $t$, we find that the particle comes to a momentary halt at $t = 3 	ext{ s}$.

Now, we substitute this time back into our position equation to find the displacement at that exact instant:

$$x = (3 - 3)^2 = 0^2 = 0 	ext{ m}$$

The particle returns exactly to the origin at the very moment it stops!

### The Elegance of the Work-Energy Theorem

Part (b) asks for the work done by the force in the first $6 	ext{ s}$.

While we could calculate the force and the displacement, the **Work-Energy Theorem** provides a much more elegant and faster route. The theorem states that the net work done on an object is equal to its change in kinetic energy:

$$W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$$

Let's calculate the initial and final velocities. At $t = 0 	ext{ s}$, the initial velocity is:

$$v_i = 2(0) - 6 = -6 	ext{ m/s}$$

At $t = 6 	ext{ s}$, the final velocity is:

$$v_f = 2(6) - 6 = 12 - 6 = 6 	ext{ m/s}$$

Substituting these values into the Work-Energy equation:

$$W = \frac{1}{2}m((6)^2 - (-6)^2)$$

$$W = \frac{1}{2}m(36 - 36) = 0 	ext{ J}$$

The total work done is zero!

### The Alternative Perspective: Force and Displacement

We can also verify this result using the fundamental definition of work, $W = F \cdot \Delta x$.

Since the velocity is $v = 2t - 6$, the acceleration is the derivative of velocity, $a = 2 	ext{ m/s}^2$. According to Newton's Second Law, the force is constant: $F = ma = 2m$.

Now, let's look at the displacement. At $t = 0$, the position is $x(0) = 9 	ext{ m}$. At $t = 6$, the position is $x(6) = (6 - 3)^2 = 9 	ext{ m}$.

The net displacement over the first $6 	ext{ s}$ is $\Delta x = 9 - 9 = 0 	ext{ m}$.

Therefore, the work done is $W = F \cdot \Delta x = 2m \cdot 0 = 0 	ext{ J}$. Both methods beautifully converge to the exact same result, showcasing the profound consistency of classical mechanics!

Visualized Solution

Analyzing the Given Equation

Isolating $x$

Expanding the Position Function

Finding the Velocity Function

Part (a): Time when Velocity is Zero

Part (a): Displacement at $t = 3 s$

Part (b): Work Done in First 6 Seconds

Calculating Initial and Final Velocities

Calculating the Work Done

The Sigma Insight: Work Done by Forces

Unveiling the Kinematics

The Moment of Rest

The Elegance of the Work-Energy Theorem

The Alternative Perspective

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