A body with mass 5 kg is acted upon by a force $\mathbf{F} = (-3\hat{i} + 4\hat{j})$ N. If its initial velocity at $t = 0$ is $\mathbf{v} = (6\hat{i} - 12\hat{j})$ m s$^{-1}$, the time at which it will just have a velocity along the $y$-axis is (a) never (b) 10 s (c) 2 s (d) 15 s

> [!TLDR] > The key idea is that the force causes constant acceleration, which changes the $x$-component of velocity linearly with time. The velocity becomes purely along the $y$-axis when the $x$-component becomes zero. Solving $v_x(t)=0$ gives $t = 10$ s. This is a straightforward application of Newton’s second law and kinematics in vector form. The force is constant, so the acceleration is constant. That means we can treat each component of motion independently — the $x$ and $y$ motions are completely decoupled. The velocity will be along the $y$-axis when its $x$-component vanishes, regardless of what the $y$-component is doing. Let’s work it through. 1. **Find the acceleration vector.** Newton’s second law: $\mathbf{F} = m \mathbf{a}$. With $m = 5$ kg and $\mathbf{F} = (-3\hat{i} + 4\hat{j})$ N, $$\mathbf{a} = \frac{\mathbf{F}}{m} = \left(-\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\right) \text{ m/s}^2.$$ 2. **Write the velocity as a function of time.** For constant acceleration, $\mathbf{v}(t) = \mathbf{v}_0 + \mathbf{a} t$. Initial velocity $\mathbf{v}_0 = 6\hat{i} - 12\hat{j}$ m/s. So $$\mathbf{v}(t) = (6\hat{i} - 12\hat{j}) + \left(-\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\right) t.$$ Group components: $$v_x(t) = 6 - \frac{3}{5}t, \quad v_y(t) = -12 + \frac{4}{5}t.$$ 3. **Set the condition for velocity along the $y$-axis.** “Just have a velocity along the $y$-axis” means the $x$-component is zero (the velocity is purely vertical). Set $v_x(t) = 0$: $$6 - \frac{3}{5}t = 0 \quad \Rightarrow \quad t = \frac{6 \times 5}{3} = 10 \text{ s}.$$ At $t = 10$ s, $v_y = -12 + \frac{4}{5}(10) = -12 + 8 = -4$ m/s, so the velocity is indeed along the negative $y$-axis. …

A body with mass 5 kg is acted upon by a force mathbfF = (-3hati + 4hatj) N. If its initial velocity at t = 0

NCERT Exemplar · Q8

\mathbf{F} = (-3\hat{i} + 4\hat{j})

(a) never

(b) 10 s

(d) 15 s

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Concept understanding — Kinematics Vector Differentiation

Kinematics Vector Differentiation

Imagine you're tracking a drone flying in the sky. At any instant, it has a position — say, 30 metres east and 40 metres north of you. That's a vector: $\vec{r} = 30\hat{i} + 40\hat{j}$ . A second later, it's moved. The question kinematics asks is: how fast is that position changing? That rate of change is velocity, and to get it, you differentiate the position vector.

But here's the key difference from school calculus: in school, you differentiated a scalar function like $y = x^2$ . Here, you're differentiating a vector function — something that has both magnitude and direction, and both can change with time.

The Intuition First

Think of a vector as an arrow. When time passes, that arrow can do two things:

It can get longer or shorter (magnitude changes).
It can rotate (direction changes).

Velocity is the total rate of change of that arrow. If the drone flies straight away from you, only the length changes. If it flies in a circle around you, only the direction changes. Most real motion does both.

So vector differentiation is just: take the derivative of each component separately, because components are independent scalars.

The Precise Statement

If a position vector is written in Cartesian coordinates as:

\vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j} + z(t)\,\hat{k}

where $\hat{i}, \hat{j}, \hat{k}$ are fixed unit vectors (they don't change direction with time), then:

\frac{d\vec{r}}{dt} = \frac{dx}{dt}\,\hat{i} + \frac{dy}{dt}\,\hat{j} + \frac{dz}{dt}\,\hat{k}

That's it. You differentiate each component function $x(t), y(t), z(t)$ exactly as you would in single-variable calculus, and the unit vectors stay put.

\frac{d}{dt}\big(f(t)\,\hat{u}\big) = \frac{df}{dt}\,\hat{u} \quad\text{(if $\hat{u}$ is constant)}

Why This Works

The derivative of a vector is defined the same way as for a scalar — as a limit:

\frac{d\vec{r}}{dt} = \lim_{\Delta t \to 0} \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}

The numerator is a vector difference. When you write $\vec{r}$ in components, the difference splits into component differences. The limit then acts on each component separately because the unit vectors are constant. So the definition forces component-wise differentiation.

A Concrete Example

A particle moves such that:

\vec{r}(t) = (3t^2)\,\hat{i} + (5\sin t)\,\hat{j} + (2e^{-t})\,\hat{k}

Its velocity is:

\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t)\,\hat{i} + (5\cos t)\,\hat{j} + (-2e^{-t})\,\hat{k}

Notice: the $x$ -component grows linearly, the $y$ -component oscillates, the $z$ -component decays. Each derivative is just the ordinary derivative of that component's function.

The One Trap: Non-Constant Unit Vectors

The rule above assumes $\hat{i}, \hat{j}, \hat{k}$ are fixed. That's true in Cartesian coordinates. But in polar coordinates, the unit vectors $\hat{r}$ and $\hat{\theta}$ rotate as the particle moves. Differentiating a vector in polar coordinates requires the product rule because the unit vectors themselves depend on time. …

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The key idea is that the force causes constant acceleration, which changes the $x$ -component of velocity linearly with time. The velocity becomes purely along the $y$ -axis when the $x$ -component becomes zero. Solving $v_x(t)=0$ gives $t = 10$ s.

This is a straightforward application of Newton’s second law and kinematics in vector form. The force is constant, so the acceleration is constant. That means we can treat each component of motion independently — the $x$ and $y$ motions are completely decoupled. The velocity will be along the $y$ -axis when its $x$ -component vanishes, regardless of what the $y$ -component is doing.

Let’s work it through.

Find the acceleration vector. Newton’s second law: $\mathbf{F} = m \mathbf{a}$ . With $m = 5$ kg and $\mathbf{F} = (-3\hat{i} + 4\hat{j})$ N,

\mathbf{a} = \frac{\mathbf{F}}{m} = \left(-\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\right) \text{ m/s}^2.

Write the velocity as a function of time. For constant acceleration, $\mathbf{v}(t) = \mathbf{v}_0 + \mathbf{a} t$ . Initial velocity $\mathbf{v}_0 = 6\hat{i} - 12\hat{j}$ m/s. So

\mathbf{v}(t) = (6\hat{i} - 12\hat{j}) + \left(-\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\right) t.

Group components:

v_x(t) = 6 - \frac{3}{5}t, \quad v_y(t) = -12 + \frac{4}{5}t.

Set the condition for velocity along the $y$ -axis. “Just have a velocity along the $y$ -axis” means the $x$ -component is zero (the velocity is purely vertical). Set $v_x(t) = 0$ :

6 - \frac{3}{5}t = 0 \quad \Rightarrow \quad t = \frac{6 \times 5}{3} = 10 \text{ s}.

At $t = 10$ s, $v_y = -12 + \frac{4}{5}(10) = -12 + 8 = -4$ m/s, so the velocity is indeed along the negative $y$ -axis. …

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Q.A body with mass 5 kg is acted upon by a force $\mathbf{F} = (-3\hat{i} + 4\hat{j})$ N. If its initial velocity at $t = 0$ is $\mathbf{v} = (6\hat{i} - 12\hat{j})$ m s $^{-1}$ , the time at which it will just have a velocity along the $y$ -axis is

Concept understanding — Kinematics Vector Differentiation

Kinematics Vector Differentiation

The Intuition First

The Precise Statement

Why This Works

A Concrete Example

The One Trap: Non-Constant Unit Vectors

Concept: Constant force ⇒ constant acceleration ⇒ independent component motion

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