Starting from the origin at time t = 0, with initial velocity 5j ms^-1, a particle moves in the xy-p

Q: Why can we just split the equation into x and y parts?

Because the x and y axes are orthogonal (perpendicular) to each other. In physics, motion along one axis does not affect motion along a perpendicular axis. This is a fundamental property of vectors, allowing us to solve complex 2D problems as two simple 1D problems.

Q: Why did we use the second equation of motion ($s = ut + 1/2 at^2$) instead of the others?

We chose this equation because it perfectly links the variables we know (initial velocity, acceleration, and displacement) with the variable we want to find (time). The first equation ($v = u + at$) requires final velocity, which we don't have. The third equation ($v^2 = u^2 + 2as$) doesn't include time, which is our primary target.

Q: Why is $u_x$ zero?

Look closely at the initial velocity given in the problem: $5\hat{j}$. The $\hat{j}$ unit vector points purely along the y-axis. Because there is no $\hat{i}$ term, it means the particle has zero initial speed in the horizontal (x) direction.

Q: What is the actual shape of the path the particle takes?

It's a parabola! If you look at the equations, $x$ is proportional to $t^2$, and $y$ is proportional to $t^2$ and $t$. If you eliminate $t$ to find the relation between $y$ and $x$ (the equation of trajectory), you will get a quadratic equation, which graphs as a parabola.

Q: Could we have solved for $y_0$ first?

No, because the y-axis equation has two unknowns: $y_0$ and $t$. The x-axis equation only has one unknown ($t$), because we were given the final x-displacement ($20\text{ m}$). You always have to start with the axis that gives you enough information to solve for the shared variable, which is time.

Question

Starting from the origin at time $t =0$, with initial velocity $5\hat{j}	ext{ ms}^{-1}$, a particle moves in the xy-plane with a constant acceleration of $(10\hat{i} + 4\hat{j})	ext{ ms}^{-2}$. At time $t$, its coordinates are $(20	ext{ m}, y_0	ext{ m})$. The values of $t$ and $y_0$ respectively, are

Options:

(A) 2 s and 18 m

(B) 5 s and 25 m

(C) 2 s and 24 m

(D) 4 s and 52 m

Sigma Percentile · Accepted Answer

### The Magic of Independent Dimensions

Imagine a particle sitting right at the origin of our coordinate system. At time $t = 0$, it is given a kick, but strictly in the y-direction. This gives us our initial velocity vector: $\mathbf{u} = 5\hat{j}	ext{ m/s}$.

However, the universe isn't letting it just travel upwards. There is a constant force acting on it, creating an acceleration of $\mathbf{a} = 10\hat{i} + 4\hat{j}	ext{ m/s}^2$. This means the particle is being pulled to the right (along the x-axis) and upwards (along the y-axis) simultaneously.

Because the acceleration is constant, we can confidently bring in our trusty kinematic equations. Specifically, the second equation of motion:

$$\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$$

The absolute superpower we have in two-dimensional kinematics is the **principle of independence of motion**. We can completely decouple the x-axis motion from the y-axis motion and solve them as two separate, highly manageable one-dimensional problems.

### Analyzing the X-Axis

Let's attack the x-axis first. Why? Because the problem gives us a massive clue: we know the final destination's x-coordinate is $20	ext{ m}$.

Looking at our initial conditions, the initial velocity has absolutely no x-component, so $u_x = 0$. The acceleration in the x-direction is simply the $\hat{i}$ component, which is $a_x = 10	ext{ m/s}^2$. Let's set up our raw equation by substituting these known values into the 1D version of our kinematic equation:

$$x = u_x t + \frac{1}{2}a_x t^2$$

$$20 = (0)t + \frac{1}{2}(10)t^2$$

Now, we execute the math. The first term vanishes entirely. Half of $10$ is $5$, leaving us with:

$$20 = 5t^2$$

Dividing both sides by $5$, we get $t^2 = 4$. Taking the positive square root (since time must be positive), we find that the particle takes exactly **$t = 2	ext{ s}$** to reach that x-coordinate.

### Conquering the Y-Axis

We've unlocked the time! Now, let's pivot to the y-axis to find that mysterious $y_0$.

We know the initial y-velocity is $u_y = 5	ext{ m/s}$, and the y-acceleration is $a_y = 4	ext{ m/s}^2$. Crucially, we now know the journey takes $2$ seconds. Let's substitute all these pieces into the y-axis displacement equation:

$$y = u_y t + \frac{1}{2}a_y t^2$$

$$y_0 = 5(2) + \frac{1}{2}(4)(2)^2$$

Let's crunch the numbers. $5$ multiplied by $2$ gives us $10$. For the second term, $2$ squared is $4$, multiplied by $4$ is $16$, and half of that is $8$.

$$y_0 = 10 + 8$$

Adding $10$ and $8$, we arrive at **$y_0 = 18	ext{ m}$**.

Bringing it all together, the time $t$ is $2	ext{ s}$, and the y-coordinate $y_0$ is $18	ext{ m}$. This perfectly matches option (a). A beautiful application of independent 1D motions to solve a 2D problem!

Select Answer:

Visualized Solution

\text{Initial Conditions}

\text{Motion in 2D}

\text{Analyzing X-axis Motion}

\text{Solving for Time } t

\text{Analyzing Y-axis Motion}

\text{Calculating } y_0

\text{Final Conclusion}

\text{The Way Forward}

The Sigma Insight: Motion in a Plane

The Magic of Independent Dimensions

Analyzing the X-Axis

Conquering the Y-Axis

Similar Questions

A particle has an initial velocity $3 \hat{i} + 4 \hat{j}$ and an acceleration of $0.4 \hat{i} + 0.3 \hat{j}$ . Its speed after $10 s$ is

The position vector of particle changes with time according to the relation $r (t) = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$ . What is the magnitude of the acceleration (in $ms^{- 2}$ ) at $t = 1$ ?

The coordinates of a moving particle at any time $t$ are given by $x = α t^{3}$ and $y = β t^{3}$ . The speed of the particle at time $t$ is given by

A particle is moving with a velocity $v = k (y \hat{i} + x \hat{j})$ , where $k$ is a constant. The general equation for its path is

A particle is moving with velocity $v = k (y \hat{i} + x \hat{j})$ , where $k$ is a constant. The general equation for its path is

The coordinates of a particle moving in a plane are given by $x (t) = a cos (pt)$ and $y (t) = b sin (pt)$ where $a, b$ ( $< a$ ) and $p$ are positive constants of appropriate dimensions. Then,

In three dimensional system, the position coordinates of a particle (in motion) are given below $x = a cos ω t, y = a sin ω t, z = aω t$ The velocity of particle will be

A particle is moving Eastwards with a velocity of $5 ms^{- 1}$ . In $10 s$ , the velocity changes to $5 ms^{- 1}$ Northwards. The average acceleration in this time is