Concept Card: Master Review & Strategy Sheet for 1D Kinematics
1. The Calculus Backbone of Kinematics:
When acceleration is non-constant, kinematic formulas cannot be applied directly; calculus definitions must be used:
- Velocity: $v(t) = \frac{dx}{dt} \iff x(t) = x_0 + \int_{0}^{t} v(t’)\,dt’$
- Acceleration (Time Form): $a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2} \iff v(t) = u + \int_{0}^{t} a(t’)\,dt’$
- Acceleration (Spatial Chain Rule Form): $a(x) = v \frac{dv}{dx} \iff \int_{u}^{v} v\,dv = \int_{x_0}^{x} a(x’)\,dx’ \implies \frac{v^2 – u^2}{2} = \int_{x_0}^{x} a(x’)\,dx’$
2. Uniformly Accelerated Motion Master Formulas ($a = \text{constant}$):
- $v = u + at$
- $s = ut + \frac{1}{2}at^2$
- $v^2 = u^2 + 2as$
- Displacement in the $n$-th second: $s_n = u + \frac{a}{2}(2n – 1)$
- Average velocity under constant acceleration: $v_{\text{avg}} = \frac{u + v}{2} \implies s = \left(\frac{u + v}{2}\right) t$
- Velocity at the geometric midpoint: $v_{\text{mid}} = \sqrt{\frac{u^2 + v^2}{2}}$
- Stopping distance under constant retardation $a$: $d_{\text{stop}} = \frac{u^2}{2a} \propto u^2$
- Two-phase motion (accelerating at $\alpha$ from rest, decelerating at $\beta$ to rest in total time $T$):
$v_{\text{max}} = \frac{\alpha\beta}{\alpha + \beta} T, \quad S_{\text{total}} = \frac{1}{2}\left(\frac{\alpha\beta}{\alpha + \beta}\right) T^2$
3. Motion Under Gravity Formulas ($a = -g$):
- Upward projection from ground: $H_{\text{max}} = \frac{u^2}{2g}, \quad t_a = t_d = \frac{u}{g}, \quad T = \frac{2u}{g}, \quad |v_{\text{landing}}| = u$
- Two-time intermediate height property: $h = \frac{1}{2}g t_1 t_2, \quad t_1 + t_2 = \frac{2u}{g}$
- Free fall from rest: $v = \sqrt{2gh}, \quad t = \sqrt{\frac{2h}{g}}, \quad y_1 : y_2 : y_3 : \dots = 1 : 3 : 5 : 7 : \dots$
- Tower projection theorem: $t_{\text{drop}} = \sqrt{t_{\text{up}} t_{\text{down}}}, \quad H = \frac{1}{2}g t_{\text{up}} t_{\text{down}}$
4. Graph Transformation Principles:
| Graph | Slope Represents | Area Under Curve Represents |
|---|---|---|
| Position-Time ($x-t$) | Instantaneous Velocity ($v = \frac{dx}{dt}$) | No direct physical meaning |
| Velocity-Time ($v-t$) | Instantaneous Acceleration ($a = \frac{dv}{dt}$) | Signed Area $= \Delta x$ (Displacement) Unsigned Area $= s$ (Total Distance) |
| Acceleration-Time ($a-t$) | Jerk ($j = \frac{da}{dt}$) | Change in Velocity ($\Delta v = v_f – v_i$) |
| Acceleration-Position ($a-x$) | Rate of change of acceleration | $\int a\,dx = \frac{v^2 – u^2}{2}$ (Specific Work / Kinetic Energy change) |
| Velocity-Position ($v-x$) | $\frac{dv}{dx}$ | Acceleration is $a = v \times (\text{Slope of } v-x)$ |
Solved Examples
Example 1 (Non-Linear Retardation $a = -k \sqrt{v}$):
A particle moving along a straight line with initial velocity $u$ experiences a deceleration proportional to the square root of its instantaneous speed: $a = -k \sqrt{v}$, where $k$ is a positive constant.
Find:
(a) The total time $T$ required for the particle to come to complete rest.
(b) The total distance $S$ covered by the particle before stopping.
Solution:
(a) Using $a = \frac{dv}{dt}$:
$\frac{dv}{dt} = -k \sqrt{v} \implies \frac{dv}{\sqrt{v}} = -k\,dt$.
Integrate from $t = 0$ ($v = u$) to $t = T$ ($v = 0$):
$\int_{u}^{0} v^{-1/2}\,dv = -k \int_{0}^{T} dt \implies [2\sqrt{v}]_{u}^{0} = -k T \implies 0 – 2\sqrt{u} = -k T$.
$T = \frac{2\sqrt{u}}{k}$.
(b) Using spatial acceleration $a = v \frac{dv}{ds}$:
$v \frac{dv}{ds} = -k \sqrt{v} \implies \frac{v}{\sqrt{v}}\,dv = -k\,ds \implies \sqrt{v}\,dv = -k\,ds$.
Integrate from $s = 0$ ($v = u$) to $s = S$ ($v = 0$):
$\int_{u}^{0} v^{1/2}\,dv = -k \int_{0}^{S} ds \implies \left[\frac{2}{3}v^{3/2}\right]_{u}^{0} = -k S \implies 0 – \frac{2}{3}u^{3/2} = -k S$.
$S = \frac{2 u^{3/2}}{3 k}$.
Example 2 (Trapezoidal $a-t$ Graph & Velocity Derivation):
A body starts from rest at $t = 0$ from the origin ($x = 0$). Its acceleration varies with time according to a trapezoidal profile:
– From $t = 0$ to $t = 2\text{ s}$, $a$ increases linearly from $0$ to $4\text{ m/s}^2$.
– From $t = 2\text{ s}$ to $t = 6\text{ s}$, $a$ remains constant at $4\text{ m/s}^2$.
– From $t = 6\text{ s}$ to $t = 8\text{ s}$, $a$ decreases linearly from $4\text{ m/s}^2$ to $0$.
(a) Determine the maximum velocity acquired by the body.
(b) Find the position $x$ of the body at $t = 8\text{ s}$.
Solution:
(a) Change in velocity $\Delta v$ is the area under the $a-t$ graph:
The $a-t$ curve is a trapezoid with base $T = 8\text{ s}$, top base $t_2 = 6 – 2 = 4\text{ s}$, and height $a_{\text{max}} = 4\text{ m/s}^2$.
$\Delta v = \text{Area} = \frac{1}{2}(\text{Base}_1 + \text{Base}_2) \times \text{Height} = \frac{1}{2}(8 + 4) \times 4 = 12 \times 2 = 24\text{ m/s}$.
Since $u = 0$, $v_{\text{max}} = 24\text{ m/s}$.
(b) To find displacement, break down into stages:
– Stage 1 ($0 \to 2\text{ s}$): $a(t) = 2t$. $v(t) = \int 2t\,dt = t^2$. $v(2) = 4\text{ m/s}$.
$x(2) = \int_{0}^{2} t^2\,dt = \left[\frac{t^3}{3}\right]_{0}^{2} = \frac{8}{3}\text{ m}$.
– Stage 2 ($2 \to 6\text{ s}$): $a = 4\text{ m/s}^2$. $v(t) = 4 + 4(t – 2)$.
Displacement $\Delta x_2 = u t’ + \frac{1}{2}a t’^2 = 4(4) + \frac{1}{2}(4)(4^2) = 16 + 32 = 48\text{ m}$.
Velocity at $t = 6\text{ s}$: $v(6) = 4 + 4(4) = 20\text{ m/s}$.
– Stage 3 ($6 \to 8\text{ s}$): $a(t’) = 4 – 2t’$ where $t’ = t – 6 \in [0, 2]$.
$v(t’) = 20 + \int_{0}^{t’} (4 – 2\tau)\,d\tau = 20 + 4t’ – t’^2$.
$\Delta x_3 = \int_{0}^{2} (20 + 4t’ – t’^2)\,dt’ = [20t’ + 2t’^2 – \frac{t’^3}{3}]_{0}^{2} = 40 + 8 – \frac{8}{3} = 48 – \frac{8}{3} = \frac{136}{3}\text{ m}$.
Total displacement: $x(8) = \frac{8}{3} + 48 + \frac{136}{3} = \frac{144}{3} + 48 = 48 + 48 = 96\text{ meters}$.
Example 3 (Simultaneous Ground Impact of Two Dropped/Projected Balls):
From the top of a $100\text{ m}$ high cliff, ball $A$ is dropped from rest at $t = 0$. At $t = 1.0\text{ s}$, ball $B$ is projected vertically downwards from the same spot with initial speed $u$. If both balls hit the ground at the exact same instant, find $u$ and the maximum separation between them before impact (take $g = 10\text{ m/s}^2$).
Solution:
1. Time taken by ball $A$ to fall $100\text{ m}$:
$H = \frac{1}{2}gt_A^2 \implies 100 = \frac{1}{2}(10)t_A^2 \implies 5 t_A^2 = 100 \implies t_A = \sqrt{20} = 2\sqrt{5}\text{ s} \approx 4.472\text{ s}$.
2. Time available for ball $B$:
$t_B = t_A – 1.0 = (2\sqrt{5} – 1)\text{ s} \approx 3.472\text{ s}$.
Equation for ball $B$:
$H = u t_B + \frac{1}{2}g t_B^2 \implies 100 = u(3.472) + 5(3.472^2) = 3.472 u + 5(12.055) = 3.472 u + 60.27$.
$3.472 u = 100 – 60.27 = 39.73 \implies u = \frac{39.73}{3.472} \approx 11.44\text{ m/s}$.
3. Maximum separation:
At $t = 1.0\text{ s}$, ball $A$ has fallen $y_A(1) = \frac{1}{2}(10)(1^2) = 5\text{ m}$ with velocity $v_A(1) = 10\text{ m/s}$.
Separation $\Delta y(t’) = (y_A – y_B)$. Since both experience acceleration $g$, relative acceleration is zero!
Relative velocity: $v_{\text{rel}} = v_A(1) – u = 10 – 11.44 = -1.44\text{ m/s}$.
Since relative velocity is negative, ball $B$ closes the gap immediately after $t = 1\text{ s}$.
Hence, the maximum separation occurs exactly at $t = 1.0\text{ s}$ when ball $B$ is released: $\Delta y_{\text{max}} = 5.0\text{ meters}$.
Example 4 (Velocity-Position Graph Analysis):
The velocity-position ($v-x$) graph of a particle moving along a straight line is a line segment connecting $(x = 0, v = v_0)$ to $(x = x_0, v = 0)$.
(a) Determine acceleration $a(x)$ as a function of position.
(b) Find the acceleration at $x = 0$ and $x = x_0$.
Solution:
(a) Equation of the straight line in the $v-x$ plane:
$v(x) = v_0 – \left(\frac{v_0}{x_0}\right)x = v_0\left(1 – \frac{x}{x_0}\right)$.
Slope of the $v-x$ graph: $\frac{dv}{dx} = -\frac{v_0}{x_0} = \text{constant}$.
Acceleration using the spatial chain rule: $a = v \frac{dv}{dx}$.
$a(x) = \left[v_0\left(1 – \frac{x}{x_0}\right)\right] \times \left(-\frac{v_0}{x_0}\right) = -\frac{v_0^2}{x_0}\left(1 – \frac{x}{x_0}\right) = \frac{v_0^2}{x_0^2}x – \frac{v_0^2}{x_0}$.
(b) At $x = 0$: $a(0) = -\frac{v_0^2}{x_0}$.
At $x = x_0$: $a(x_0) = 0$.
Insight: The $a-x$ graph is a straight line with positive slope $\frac{v_0^2}{x_0^2}$ starting from negative acceleration $-\frac{v_0^2}{x_0}$ and reaching zero at $x = x_0$.
Worksheet: Unit Test (10 Questions)
Problem 1 (JEE Main – Single Correct):
A particle moves along a straight line such that its displacement $x$ at time $t$ is related by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero is:
(A) $0\text{ m}$
(B) $3\text{ m}$
(C) $9\text{ m}$
(D) $12\text{ m}$
Problem 2 (JEE Main – Single Correct):
The acceleration $a$ of a particle moving along the $x$-axis varies with position as $a = -4x\text{ m/s}^2$. If the particle starts from rest at $x = 2\text{ m}$, its velocity when passing through the origin ($x = 0$) is:
(A) $2\text{ m/s}$
(B) $4\text{ m/s}$
(C) $8\text{ m/s}$
(D) $16\text{ m/s}$
Problem 3 (JEE Main – Single Correct):
A hot-air balloon is ascending vertically with a constant speed of $12\text{ m/s}$. When it is at a height of $65\text{ m}$ above the ground, a packet is dropped from it. Taking $g = 10\text{ m/s}^2$, the time taken by the packet to reach the ground is:
(A) $5\text{ s}$
(B) $6.5\text{ s}$
(C) $4\text{ s}$
(D) $3\text{ s}$
Problem 4 (JEE Main – Single Correct):
If the momentum of an automobile is increased by $100\%$ while the maximum braking force remains unchanged, its minimum stopping distance will increase by:
(A) $100\%$
(B) $200\%$
(C) $300\%$
(D) $400\%$
Problem 5 (JEE Main – Single Correct):
The velocity of a particle is given by $v(t) = 3t^2 – 6t\text{ m/s}$. The distance traveled by the particle in the interval $t = 0$ to $t = 3\text{ s}$ is:
(A) $0\text{ m}$
(B) $4\text{ m}$
(C) $8\text{ m}$
(D) $12\text{ m}$
Problem 6 (JEE Advanced – One or More Correct):
An acceleration-position ($a-x$) graph of a particle starting from rest at $x = 0$ is a triangle with base from $x = 0$ to $x = 8\text{ m}$ and peak acceleration $6\text{ m/s}^2$ at $x = 4\text{ m}$. Which of the following statements is/are correct?
(A) The area under the $a-x$ curve equals $\frac{v^2 – u^2}{2}$.
(B) The area under the $a-x$ curve is $24\text{ m}^2/\text{s}^2$.
(C) The speed of the particle at $x = 8\text{ m}$ is $4\sqrt{3}\text{ m/s}$.
(D) The speed of the particle at $x = 4\text{ m}$ is $2\sqrt{6}\text{ m/s}$.
Problem 7 (JEE Advanced – One or More Correct):
A ball is dropped from a high tower of height $H$. It covers a distance $h_1$ in the first third of the total fall time, and $h_2$ in the final third of the total fall time. Which statements are correct?
(A) $h_1 = \frac{1}{9}H$
(B) $h_2 = \frac{5}{9}H$
(C) The ratio $\frac{h_2}{h_1} = 5$
(D) The speed at the end of the second third of total time is $\frac{2}{3}v_{\text{ground}}$
Problem 8 (JEE Advanced – One or More Correct):
Two particles $A$ and $B$ move along the $x$-axis. Their position equations are $x_A(t) = 3t^2 – 4t$ and $x_B(t) = t^2 + 2t$. Which statements are TRUE?
(A) They have equal velocity at $t = 1.5\text{ s}$.
(B) Their separation is minimum at $t = 1.5\text{ s}$.
(C) They cross each other at $t = 0$ and $t = 3\text{ s}$.
(D) Relative acceleration between them is constant at $4\text{ m/s}^2$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A particle moves along a straight line with acceleration $a(t) = 6t – 4\text{ m/s}^2$. At $t = 0$, its position is $x(0) = 2\text{ m}$ and velocity is $v(0) = 3\text{ m/s}$. Find the position of the particle in meters at $t = 2\text{ seconds}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A train travels between two stations $600\text{ m}$ apart. It accelerates from rest at rate $\alpha = 2.0\text{ m/s}^2$ and then immediately decelerates to rest at rate $\beta = 4.0\text{ m/s}^2$. The maximum speed reached by the train in $\text{m/s}$ is $V_{\text{max}}$. Find the integer value of $V_{\text{max}}$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (A) | 4. (C) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 8 | 10. 40
Solution 1:
$t = \sqrt{x} + 3 \implies \sqrt{x} = t – 3 \implies x(t) = (t – 3)^2$.
$v(t) = \frac{dx}{dt} = 2(t – 3)$.
$v = 0 \implies t = 3\text{ s}$.
At $t = 3\text{ s}$: $x(3) = (3 – 3)^2 = 0\text{ m}$.
Correct Answer: (A)
Solution 2:
$a = v \frac{dv}{dx} = -4x \implies v\,dv = -4x\,dx$.
Integrating from $x = 2$ ($v = 0$) to $x = 0$ ($v$):
$\int_{0}^{v} v\,dv = -4 \int_{2}^{0} x\,dx \implies \frac{v^2}{2} = -4 \left[\frac{x^2}{2}\right]_{2}^{0} = -2 (0 – 4) = 8$.
$v^2 = 16 \implies v = 4\text{ m/s}$.
Correct Answer: (B)
Solution 3:
$y(t) = ut – \frac{1}{2}gt^2$. Ground is at $y = -65\text{ m}$.
$-65 = 12t – 5t^2 \implies 5t^2 – 12t – 65 = 0$.
Factoring: $(t – 5)(5t + 13) = 0 \implies t = 5\text{ seconds}$.
Correct Answer: (A)
Solution 4:
Momentum doubles ($p’ = 2p \implies v’ = 2v$).
Stopping distance: $d_{\text{stop}} = \frac{v^2}{2a} \propto v^2$.
New distance: $d’ = (2v)^2 / (2a) = 4 d_{\text{stop}}$.
Percentage increase: $\frac{4d – d}{d} \times 100\% = 300\%$.
Correct Answer: (C)
Solution 5:
$v(t) = 3t^2 – 6t = 3t(t – 2) = 0$ at $t = 2\text{ s}$ (turning point).
$x(t) = \int (3t^2 – 6t)\,dt = t^3 – 3t^2$.
$x(0) = 0$, $x(2) = 8 – 12 = -4\text{ m}$, $x(3) = 27 – 27 = 0\text{ m}$.
Total distance: $|x(2) – x(0)| + |x(3) – x(2)| = |-4 – 0| + |0 – (-4)| = 4 + 4 = 8\text{ m}$.
Correct Answer: (C)
Solution 6:
– Area of triangle: $\frac{1}{2} \times 8 \times 6 = 24\text{ m}^2/\text{s}^2$ (B is True).
– Work-energy identity: $\int a\,dx = \frac{v^2 – u^2}{2} = 24 \implies v^2 = 48 \implies v(8) = \sqrt{48} = 4\sqrt{3}\text{ m/s}$ (A, C are True).
– At $x = 4\text{ m}$, area of left triangle is $\frac{1}{2} \times 4 \times 6 = 12 \implies \frac{v^2}{2} = 12 \implies v(4) = \sqrt{24} = 2\sqrt{6}\text{ m/s}$ (D is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
Let total time be $T$. $H = \frac{1}{2}g T^2$.
– In first third ($T/3$): $h_1 = \frac{1}{2}g(T/3)^2 = \frac{1}{9}H$ (A is True).
– In first two-thirds ($2T/3$): $y = \frac{1}{2}g(2T/3)^2 = \frac{4}{9}H$.
– In final third: $h_2 = H – \frac{4}{9}H = \frac{5}{9}H$ (B is True).
– Ratio $\frac{h_2}{h_1} = \frac{5/9}{1/9} = 5$ (C is True).
– Speed at $t = \frac{2}{3}T$ is $v = g\left(\frac{2}{3}T\right) = \frac{2}{3}(gT) = \frac{2}{3}v_{\text{ground}}$ (D is True).
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– Relative position: $x_{\text{rel}} = x_B – x_A = (t^2 + 2t) – (3t^2 – 4t) = -2t^2 + 6t$.
– Crossings when $x_{\text{rel}} = 0 \implies 2t(3 – t) = 0 \implies t = 0$ and $t = 3\text{ s}$ (C is True).
– Separation $\Delta x = 6t – 2t^2$. Max/extremum when $\frac{d}{dt}(6t – 2t^2) = 6 – 4t = 0 \implies t = 1.5\text{ s}$ (A and B are True).
– Relative acceleration: $a_A = 6, a_B = 2 \implies a_{\text{rel}} = 6 – 2 = 4\text{ m/s}^2$ (D is True).
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 9:
$a(t) = 6t – 4$.
$v(t) = v(0) + \int_{0}^{t} (6\tau – 4)\,d\tau = 3 + 3t^2 – 4t$.
$x(t) = x(0) + \int_{0}^{t} (3 + 3\tau^2 – 4\tau)\,d\tau = 2 + 3t + t^3 – 2t^2$.
At $t = 2\text{ s}$:
$x(2) = 2 + 3(2) + 2^3 – 2(2^2) = 2 + 6 + 8 – 8 = 8\text{ meters}$.
Correct Answer: 8
Solution 10:
For two-stage motion from rest to rest:
$S = \frac{1}{2}\left(\frac{\alpha + \beta}{\alpha\beta}\right) v_{\text{max}}^2$.
$600 = \frac{1}{2}\left(\frac{2 + 4}{2 \times 4}\right) v_{\text{max}}^2 = \frac{1}{2}\left(\frac{6}{8}\right) v_{\text{max}}^2 = \frac{3}{8} v_{\text{max}}^2$.
$v_{\text{max}}^2 = \frac{600 \times 8}{3} = 200 \times 8 = 1600$.
$v_{\text{max}} = \sqrt{1600} = 40\text{ m/s}$.
Correct Answer: 40