Motion in a plane can be described using Cartesian coordinates, $\vec{A} = A_x\hat{i} + A_y\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along the $x$ and $y$ directions. It can equally be described using plane polar coordinates, $\vec{A} = A_r\hat{r} + A_\theta\hat{\theta}$, where $$\hat{r} = \frac{\vec{r}}{r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}, \qquad \hat{\theta} = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}$$ are unit vectors along the directions in which $r$ and $\theta$ increase, respectively. (a) Express $\hat{i}$ and $\hat{j}$ in terms of $\hat{r}$ and $\hat{\theta}$. (b) Show that $\hat{r}$ and $\hat{\theta}$ are both unit vectors and are perpendicular to each other. (c) Show that $\dfrac{d}{dt}(\hat{r}) = \omega\hat{\theta}$ and $\dfrac{d}{dt}(\hat{\theta}) = -\omega\hat{r}$, where $\omega = \dfrac{d\theta}{dt}$. (d) For a particle moving along a spiral given by $\vec{r} = a\theta\,\hat{r}$ with $a = 1$ (unit), find the dimensions of $a$. (e) Find the velocity and acceleration in polar-vector form for the particle moving along the spiral in (d).

Motion in a plane can be described using Cartesian coordinates, vecA = Aₓhati + Ayhatj, where hati and hatj ar

NCERT Exemplar · Q36

\vec{A} = A_x\hat{i} + A_y\hat{j}

(a) Express

\hat{i}

and

\hat{j}

in terms of

\hat{r}

and

\hat{\theta}

(b) Show that

\hat{r}

and

\hat{\theta}

are both unit vectors and are perpendicular to each other.

\dfrac{d}{dt}(\hat{r}) = \omega\hat{\theta}

and

\dfrac{d}{dt}(\hat{\theta}) = -\omega\hat{r}

, where

\omega = \dfrac{d\theta}{dt}

(d) For a particle moving along a spiral given by

\vec{r} = a\theta\,\hat{r}

with

a = 1

(unit), find the dimensions of

a

(e) Find the velocity and acceleration in polar-vector form for the particle moving along the spiral in (d).

Andhra Pradesh BieapLong· 5mImportance★★★★★est

99% · 67/68 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Parts (a)-(c) are algebra with the given definitions of $\hat{r}$ and $\hat{\theta}$ : invert them, check magnitudes and the dot product, and differentiate. In (d) the relation $r=a\theta$ with dimensionless $\theta$ forces $a$ to carry the dimension of length. In (e) differentiate $\vec{r}=a\theta\hat{r}$ using the results of (c).

(a) $\hat{i}$ and $\hat{j}$ in terms of $\hat{r},\hat{\theta}$

Given $\hat{r}=\cos\theta\,\hat{i}+\sin\theta\,\hat{j}$ and $\hat{\theta}=-\sin\theta\,\hat{i}+\cos\theta\,\hat{j}$ . Form the combinations:

\cos\theta\,\hat{r}-\sin\theta\,\hat{\theta} = (\cos^2\theta+\sin^2\theta)\hat{i} = \hat{i},

\sin\theta\,\hat{r}+\cos\theta\,\hat{\theta} = (\sin^2\theta+\cos^2\theta)\hat{j} = \hat{j}.

So $\hat{i}=\cos\theta\,\hat{r}-\sin\theta\,\hat{\theta}$ and $\hat{j}=\sin\theta\,\hat{r}+\cos\theta\,\hat{\theta}$ .

(b) Unit vectors and perpendicularity

|\hat{r}|^2=\cos^2\theta+\sin^2\theta=1,\qquad |\hat{\theta}|^2=\sin^2\theta+\cos^2\theta=1,

so both are unit vectors. Their dot product

\hat{r}\cdot\hat{\theta} = (\cos\theta)(-\sin\theta)+(\sin\theta)(\cos\theta)=0,

so they are perpendicular.

(c) Time derivatives

With $\omega=d\theta/dt$ ,

\frac{d\hat{r}}{dt} = (-\sin\theta\,\hat{i}+\cos\theta\,\hat{j})\frac{d\theta}{dt} = \omega\,\hat{\theta},

\frac{d\hat{\theta}}{dt} = (-\cos\theta\,\hat{i}-\sin\theta\,\hat{j})\frac{d\theta}{dt} = -\omega(\cos\theta\,\hat{i}+\sin\theta\,\hat{j}) = -\omega\,\hat{r}.

(d) Dimensions of $a$

For the spiral, the position vector is $\vec{r}=a\theta\,\hat{r}$ , so its magnitude is $r=a\theta$ . Here $r$ is a length, dimension $[\text{L}]$ , while $\theta$ (an angle in radians) is dimensionless. Hence

[a] = \frac{[r]}{[\theta]} = [\text{L}] = \text{M}^0\text{L}^1\text{T}^0,

so $a$ has the dimension of length (it is measured in metres).

(e) Velocity and acceleration on the spiral

Start from $\vec{r}=a\theta\,\hat{r}$ . Differentiate, using $\dot{\hat{r}}=\dot\theta\,\hat{\theta}$ :

\vec{v}=\frac{d}{dt}(a\theta\,\hat{r}) = a\dot\theta\,\hat{r} + a\theta\,\dot{\hat{r}} = a\dot\theta\,\hat{r} + a\theta\dot\theta\,\hat{\theta} = a\dot\theta\,(\hat{r}+\theta\,\hat{\theta}).

Differentiate again, using $\dot{\hat{r}}=\dot\theta\,\hat{\theta}$ and $\dot{\hat{\theta}}=-\dot\theta\,\hat{r}$ : …

Start your 14-day free trial to unlock the full solution →

Concept: Treat $(\hat r,\hat\theta)$ as a Rotated Copy of $(\hat\imath,\hat\jmath)$ , Using Rotation-Matrix and Rotating-Frame Facts

Method: Rotation Matrix for (a)-(b), the $\vec\omega\times\hat e$ Rule for (c), and the General Polar Velocity/Acceleration Formulas for (e)

Instead of inverting and differentiating the given sine/cosine expressions component by component, this method recognizes $(\hat r,\hat\theta)$ as the ordinary axes $(\hat\imath,\hat\jmath)$ rotated by angle $\theta$ , and applies three standard facts about rotations that make parts (a), (b), and (c) nearly automatic.

Steps

(a) Recognize the given definitions as a rotation matrix: $\begin{pmatrix}\hat r\\\hat\theta\end{pmatrix} = \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}\hat\imath\\\hat\jmath\end{pmatrix}$ . A rotation matrix's inverse is simply the rotation by the opposite angle (equivalently, its transpose, since rotation matrices are orthogonal):

\hat\imath = \cos\theta\,\hat r - \sin\theta\,\hat\theta, \qquad \hat\jmath = \sin\theta\,\hat r + \cos\theta\,\hat\theta.

(b) Unit length and perpendicularity follow automatically from "this is a rotation." A rotation matrix is orthogonal by construction (it preserves lengths and angles) -- so $\hat r,\hat\theta$ being unit vectors and mutually perpendicular doesn't need a separate dot-product calculation; it's guaranteed by rotating the orthonormal pair $(\hat\imath,\hat\jmath)$ .
(c) Use the standard rotating-unit-vector rule $\dot{\hat e} = \vec\omega\times\hat e$ (with $\vec\omega=\dot\theta\,\hat z$ for planar rotation), instead of differentiating $\cos\theta,\sin\theta$ directly:

\dot{\hat r} = \dot\theta\,\hat z\times\hat r = \dot\theta\,\hat\theta \;(=\omega\hat\theta), \qquad \dot{\hat\theta} = \dot\theta\,\hat z\times\hat\theta = -\dot\theta\,\hat r\;(=-\omega\hat r),

using the standard planar identities $\hat z\times\hat r=\hat\theta$ and $\hat z\times\hat\theta=-\hat r$ (true for any angle $\theta$ , since $\hat r,\hat\theta,\hat z$ always form a right-handed triad).

(d) Dimensional consistency of $r=a\theta$ : $\theta$ (radians) is dimensionless, $r$ has dimension $[\text L]$ , so $[a]=[\text L]$ -- unchanged from the direct method.
(e) Apply the general polar velocity/acceleration formulas (which follow from the product rule combined with the rotating-vector rule in step 3, and are worth memorizing once for any polar-coordinate problem):

\vec v = \dot r\,\hat r + r\dot\theta\,\hat\theta, \qquad \vec a = (\ddot r - r\dot\theta^2)\,\hat r + (r\ddot\theta+2\dot r\dot\theta)\,\hat\theta.

Substitute $r=a\theta$ , $\dot r=a\dot\theta$ , $\ddot r=a\ddot\theta$ : …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$2\hat{i}+3\hat{j}\ ms^{-1}$
›Reveal solutionSolution
Straightforward kinematics with vectors: add the acceleration's contribution to each component of velocity separately, then find the magnitude.

Concept and Intuition

Since acceleration is constant here, each Cartesian component of velocity evolves independently and linearly in time: $v_x(t) = v_{0x}+a_x t$ and $v_y(t) = v_{0y}+a_y t$ . There's no need to find a resultant acceleration vector's magnitude/direction first — just apply the equation component-by-component and combine at the end.

Step-by-Step Solution

Given $\vec v_0 = 2\hat i + 3\hat j\ ms^{-1}$ and $\vec a = 0.8\hat i + 0.6\hat j\ ms^{-2}$ , and $t=5\,s$ .

$x$ -component: $v_x = 2 + 0.8(5) = 2+4 = 6\ ms^{-1}$ .

$y$ -component: $v_y = 3 + 0.6(5) = 3+3 = 6\ ms^{-1}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$V = 2\sqrt{x}\ \text{ms}^{-1}$
›Reveal solutionSolution
Using $a = v\,dv/dx$ on $v = 2\sqrt{x}$ gives a constant acceleration of

$2\ \text{ms}^{-2}$ . Answer: (A).

Concept and Intuition

When velocity is given as a function of position (not time), the cleanest way

to get acceleration is the identity $a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v\dfrac{dv}{dx}$ . This avoids needing to first invert

$v(x)$ into $x(t)$ .

Step-by-Step Solution

Given $v = 2\sqrt{x} = 2x^{1/2}$ .

Differentiate w.r.t. $x$ : $\dfrac{dv}{dx} = 2\cdot\dfrac{1}{2}x^{-1/2} = \dfrac{1}{\sqrt{x}}$ .

Apply $a = v\dfrac{dv}{dx} = \left(2\sqrt{x}\right)\cdot \left(\dfrac{1}{\sqrt{x}}\right) = 2$ .

The $\sqrt{x}$ terms cancel exactly, so acceleration is the constant …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$30°$
›Reveal solutionSolution
Acceleration equals the slope of the v–t graph, so $a_1:a_2=\tan30°:\tan45°=1:\sqrt3$ .

Concept and Intuition

On a velocity-time graph, the instantaneous acceleration is $\dfrac{dv}{dt}$ , which is exactly the slope of the graph — and slope equals the tangent of the angle the line makes with the time axis.

Step-by-Step Solution

$a_1=\tan30°=\dfrac{1}{\sqrt3}$

$a_2=\tan45°=1$

Ratio:

$\frac{a_1}{a_2}=\frac{1/\sqrt3}{1}=\frac{1}{\sqrt3}$

So $a_1:a_2=1:\sqrt3$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$\vec{E} = E_0\hat{i}$
›Reveal solutionSolution
The field only accelerates the particle along $x$ ; the $y$ -velocity stays fixed at $v_0$ , so the target speed condition fixes $v_x=v_0/2$ , giving $t=\dfrac{mv_0}{2qE_0}$ .

Concept and Intuition

The electric field acts only along $\hat i$ , so it produces a constant acceleration $a=qE_0/m$ purely in the $x$ -direction — exactly like projectile motion, where the $y$ -component of velocity is completely unaffected by a force acting only along $x$ . The particle's speed at any time is found from both components combined via Pythagoras.

Step-by-Step Solution

Acceleration: $a=\dfrac{qE_0}{m}$ , directed along $x$ .

Velocity components at time $t$ : $v_x(t)=at=\dfrac{qE_0}{m}t$ , and $v_y(t)=v_0$ (unchanged, no force along $y$ ).

Required speed: $v=\dfrac{\sqrt5}{2}v_0 \Rightarrow v^2=v_x^2+v_y^2=\dfrac54v_0^2$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$v = at - bt^2$
›Reveal solutionSolution
Differentiate the given velocity to get acceleration, then find when it vanishes; $t=a/2b$ .

Concept and Intuition

Acceleration is the instantaneous rate of change of velocity. Given $v(t)$ as a function of time, $a(t)=dv/dt$ . Setting $a(t)=0$ finds the instant the velocity is momentarily stationary (an extremum of $v$ ).

Step-by-Step Solution

$v = at - bt^2$ .

$a_{acc} = \dfrac{dv}{dt} = a - 2bt$ .

Set $a_{acc}=0$ : $a - 2bt = 0 \Rightarrow t = \dfrac{a}{2b}$ .

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$^{-1}$
›Reveal solutionSolution
A velocity-vs-position relation is turned into acceleration using $a = v\,dv/dx$ . Answer: (C) 100 ms $^{-2}$ .

Concept and Intuition

When velocity is given as a function of position rather than time, the direct route to acceleration is the chain rule: $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ . This is the standard trick for "velocity increases so much per metre" problems — you never need to know how it depends on time at all.

Step-by-Step Solution

The statement "velocity is increasing at 5 ms $^{-1}$ per metre" means $\dfrac{dv}{dx} = 5\ \text{s}^{-1}$ (constant).

Use $a = v\dfrac{dv}{dx}$ .

At the instant $v = 20\ \text{ms}^{-1}$ : $a = 20 \times 5 = 100\ \text{ms}^{-2}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$t=0$
›Reveal solutionSolution
Matching the parabolic path $y=bx^2$ to the kinematic equations of constant-velocity $x$ -motion and constant-acceleration $y$ -motion gives $v_x=\sqrt{a/2b}$ .

Concept and Intuition

This is projectile-like motion: since acceleration acts only along $y$ , the $x$ -motion is uniform (constant velocity), while $y$ -motion starts from rest and accelerates uniformly. The given trajectory equation $y=bx^2$ must be consistent with both these kinematic equations at every instant $t$ , which pins down $v_x$ in terms of $a$ and $b$ .

Step-by-Step Solution

Since there's no acceleration along $x$ : $x=v_x t$ (constant velocity $v_x$ ).

Since acceleration $a$ acts along $y$ starting from rest at the origin: $y=\dfrac12 a t^2$ .

The trajectory is given as $y=bx^2$ . Substitute $x=v_xt$ : $y=b(v_xt)^2=bv_x^2t^2.$ …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$x = 2t^2 - t$
›Reveal solutionSolution
Differentiate $x(t)$ and $y(t)$ to get the velocity components, then combine them via the Pythagorean theorem to get the speed. Answer: (D) 5 m/s.

Concept and Intuition

For 2D motion described by position components $x(t)$ and $y(t)$ , the instantaneous velocity components are obtained by differentiating each with respect to time: $v_x = dx/dt$ , $v_y = dy/dt$ . The magnitude (speed) of the velocity vector is then found via $|\vec v| = \sqrt{v_x^2+v_y^2}$ , exactly as with any 2D vector.

Step-by-Step Solution

Given: $x = 2t^2 - t$ , so $v_x = \dfrac{dx}{dt} = 4t - 1$ .

Given: $y = 4t^2 - 4t$ , so $v_y = \dfrac{dy}{dt} = 8t - 4$ .

Evaluate at $t = 1\ \mathrm{s}$ : $v_x = 4(1)-1 = 3\ \mathrm{m/s}$ ; $v_y = 8(1)-4 = 4\ \mathrm{m/s}$ .

Compute the speed: $|\vec v| = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\ \mathrm{m/s}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$1:\sqrt{3}$
›Reveal solutionSolution

On this swapped-axis plot ( $t$ vertical, $s$ horizontal), $v=\frac{\Delta s}{\Delta t}$ : for A ( $45^\circ$ to the $s$ -axis) $v_A=1$ ; for B ( $30^\circ$ to the $t$ -axis) $v_B=\tan30^\circ=\frac{1}{\sqrt3}$ ; so $v_A:v_B=\sqrt3:1$ -> option (C).

The axes are unusual: time $t$ is on the vertical axis and displacement $s$ on the horizontal axis. Velocity is still $v=\dfrac{\Delta s}{\Delta t}$ = (horizontal change) $/$ (vertical change).

Body A - line at $45^\circ$ to the horizontal $s$ -axis. Measuring from the horizontal, $\tan45^\circ=\dfrac{\Delta t}{\Delta s}=1$ , so

$v_A=\frac{\Delta s}{\Delta t}=1.$ …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$t$
›Reveal solutionSolution
Differentiating $t=\alpha x^2+\beta x$ twice (via $v=dx/dt$ and $a=v\,dv/dx$ ) gives the deceleration as $2\alpha v^3$ — (D).

Concept and Intuition

When $t$ is given as a function of $x$ (rather than the usual $x(t)$ ), it's often easiest to first get $v=dx/dt$ as a function of $x$ by inverting $dt/dx$ , and then use the identity $a=v\dfrac{dv}{dx}$ to find acceleration without ever needing an explicit $x(t)$ .

Step-by-Step Solution

$t=\alpha x^2+\beta x\Rightarrow\dfrac{dt}{dx}=2\alpha x+\beta$ .

Since $v=\dfrac{dx}{dt}$ , we have $v=\dfrac{1}{dt/dx}=\dfrac1{2\alpha x+\beta}$ .

Differentiate $v$ with respect to $x$ : $\dfrac{dv}{dx}=-\dfrac{2\alpha}{(2\alpha x+\beta)^2}=-2\alpha v^2$ (recognizing $(2\alpha x+\beta)^{-2}=v^2$ ).

Acceleration: $a=v\dfrac{dv}{dx}=v\cdot(-2\alpha v^2)=-2\alpha v^3$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
$4s = 5t^{2} + 8t$
›Reveal solutionSolution
A kinematics differentiation problem: acceleration is the second derivative of position with respect to time. Answer: (B) $2.5\ \mathrm{ms^{-2}}$ .

Concept and Intuition

Given any position–time relation $s(t)$ , velocity is $v = \dfrac{ds}{dt}$ and acceleration is $a = \dfrac{dv}{dt}$ . Since the given relation is quadratic in $t$ , the motion is uniformly accelerated (constant acceleration), so differentiating twice gives a time-independent constant.

Step-by-Step Solution

Rewrite $4s = 5t^{2}+8t$ as $s = \dfrac{5}{4}t^{2} + 2t$ .

Differentiate once for velocity: $v = \dfrac{ds}{dt} = \dfrac{5}{2}t + 2$ .

Differentiate again for acceleration: $a = \dfrac{dv}{dt} = \dfrac{5}{2} = 2.5\ \mathrm{ms^{-2}}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$x(t) = \alpha t e^{-t/\tau}$
›Reveal solutionSolution
Velocity is $v = dx/dt$ . Differentiating the given product-of-exponential expression and plugging in $t = 2$ s gives $v = -1/e^2\ \mathrm{m\,s^{-1}}$ .

Concept and Intuition

Whenever position is given as an explicit function of time, velocity is simply its time derivative. Here $x(t) = \alpha t e^{-t/\tau}$ is a product of $t$ and a decaying exponential, so the product rule is required.

Step-by-Step Solution

Write $x(t) = \alpha t\, e^{-t/\tau}$ .

Differentiate using the product rule:

$v(t) = \frac{dx}{dt} = \alpha e^{-t/\tau} + \alpha t \left(-\frac{1}{\tau}\right) e^{-t/\tau} = \alpha e^{-t/\tau}\left(1 - \frac{t}{\tau}\right)$

Substitute $\alpha = 1\ \mathrm{m\,s^{-1}}$ , $\tau = 1\ \mathrm{s}$ , $t = 2\ \mathrm{s}$ : …
Start your 14-day free trial to see the full solution →

←Previous questionQ35 — A cricket fielder can throw the cricket ball with a speed vo. If he th…→Next questionQ37 — A man wants to travel from corner A of a square field to the diagonall…

(a) i^\hat{i}i^ and j^\hat{j}j^​ in terms of r^,θ^\hat{r},\hat{\theta}r^,θ^

(b) Unit vectors and perpendicularity

(c) Time derivatives

(d) Dimensions of aaa

(e) Velocity and acceleration on the spiral

🎓Unlock everything free for 14 days

(a) $\hat{i}$ and $\hat{j}$ in terms of $\hat{r},\hat{\theta}$

(d) Dimensions of $a$

Unlock everything free for 14 days