Concept Card: Third Equation of Motion & Kinematic Problem Solving
1. Mathematical Derivation of the Third Kinematic Equation ($v^2 = u^2 + 2as$):
- Calculus Derivation via Differential Chain Rule:
Acceleration is defined as the time derivative of velocity: $a = \frac{dv}{dt}$.
Applying the chain rule of differentiation with respect to spatial displacement $s$:
$a = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$
Separating variables:
$v\,dv = a\,ds$
For uniformly accelerated motion ($a = \text{constant}$), integrate both sides from initial speed $u$ (at $s = 0$) to final speed $v$ (at displacement $s$):
$\int_{u}^{v} v\,dv = \int_{0}^{s} a\,ds \implies \left[\frac{v^2}{2}\right]_{u}^{v} = a [s]_{0}^{s} \implies \frac{v^2 – u^2}{2} = as \implies v^2 = u^2 + 2as$. - Algebraic Derivation via Elimination of Time $t$:
From the first kinematic equation: $v = u + at \implies t = \frac{v – u}{a}$.
Substitute $t$ into the second kinematic equation $s = ut + \frac{1}{2}at^2$:
$s = u\left(\frac{v – u}{a}\right) + \frac{1}{2}a\left(\frac{v – u}{a}\right)^2 = \frac{2u(v – u) + (v – u)^2}{2a} = \frac{v^2 – u^2}{2a} \implies v^2 = u^2 + 2as$.
2. Connection to the Work-Kinetic Energy Theorem:
Multiply the equation $v^2 – u^2 = 2as$ by $\frac{1}{2}m$ (where $m$ is object mass):
$\frac{1}{2}mv^2 – \frac{1}{2}mu^2 = (ma)s = F_{\text{net}} s = W_{\text{net}}$
Thus, the third kinematic equation is fundamentally the one-dimensional kinematic counterpart of the Work-Energy Theorem for constant net force.
3. High-Yield Strategic Formulas for JEE:
- Velocity at the Spatial Midpoint:
If a particle with constant acceleration moves from point $A$ (speed $u$) to point $B$ (speed $v$), the speed at the exact geometric midpoint of the path is:
$v_{\text{mid}} = \sqrt{\frac{u^2 + v^2}{2}}$
(Notice that this result is completely independent of the distance $s$ and acceleration $a$). - Stopping Distance & Stopping Time:
Under constant braking deceleration $a$, setting final speed $v = 0$ yields:
$d_{\text{stop}} = \frac{u^2}{2a} \implies d_{\text{stop}} \propto u^2$
$t_{\text{stop}} = \frac{u}{a} \implies t_{\text{stop}} \propto u$
(Doubling initial speed quadruples the required stopping distance!). - Average Velocity under Constant Acceleration:
$v_{\text{avg}} = \frac{u + v}{2} \implies s = \left(\frac{u + v}{2}\right) t$ - Collision Avoidance Criterion (Relative Motion):
If Vehicle 1 (initial speed $u_1$, braking deceleration $a_1$) is trailing Vehicle 2 (moving forward at constant speed $u_2$) with initial separation $d$, collision is avoided if:
$\frac{(u_1 – u_2)^2}{2 a_1} \le d$
Solved Examples
Example 1 (Train Crossing a Signal Post & Midpoint Speed):
A train of length $L$ passes an observer standing beside the track with uniform acceleration. The front locomotive passes the observer with speed $u = 36\text{ km/h}$ ($10\text{ m/s}$) and the rear coach passes with speed $v = 72\text{ km/h}$ ($20\text{ m/s}$).
(a) Find the speed with which the exact midpoint of the train passes the observer.
(b) If the train has $10$ identical coaches, find the speed with which the front of the $6$-th coach passes the observer.
Solution:
(a) Let the length of the train be $L$.
For the complete passage of the train (from front to rear):
$v^2 = u^2 + 2aL \implies 2aL = v^2 – u^2 = 20^2 – 10^2 = 400 – 100 = 300\text{ m}^2/\text{s}^2$.
When the midpoint passes the observer, the train has traveled a distance $s = \frac{L}{2}$:
$v_{\text{mid}}^2 = u^2 + 2a\left(\frac{L}{2}\right) = u^2 + aL = u^2 + \frac{v^2 – u^2}{2} = \frac{u^2 + v^2}{2}$.
$v_{\text{mid}}^2 = \frac{10^2 + 20^2}{2} = \frac{100 + 400}{2} = 250\text{ m}^2/\text{s}^2$.
$v_{\text{mid}} = \sqrt{250} = 5\sqrt{10}\text{ m/s} \approx 15.81\text{ m/s}$ ($56.9\text{ km/h}$).
(b) The front of the $6$-th coach passing the observer means exactly $5$ coaches have crossed, which corresponds to distance $s = \frac{5}{10}L = \frac{L}{2}$.
Hence, its speed is identically $v_6 = v_{\text{mid}} = 5\sqrt{10}\text{ m/s} \approx 15.81\text{ m/s}$.
Example 2 (Bullet Penetrating a Wooden Block & Retardation):
A bullet moving with velocity $u$ strikes a fixed wooden plank and loses $\frac{1}{n}$-th of its velocity after penetrating a distance $x$. Assuming uniform resistive deceleration, determine the additional distance $d$ the bullet penetrates before coming to complete rest.
Solution:
Phase 1: Initial velocity $= u$, final velocity $= u – \frac{u}{n} = u\left(\frac{n – 1}{n}\right)$, distance $= x$.
$v^2 = u^2 – 2ax \implies \left[u\frac{n – 1}{n}\right]^2 = u^2 – 2ax$
$2ax = u^2 \left[1 – \left(\frac{n – 1}{n}\right)^2\right] = u^2 \left[\frac{n^2 – (n^2 – 2n + 1)}{n^2}\right] = u^2 \frac{2n – 1}{n^2}$.
Phase 2: From speed $u\left(\frac{n – 1}{n}\right)$ to $0$ over additional distance $d$:
$0 = \left[u\frac{n – 1}{n}\right]^2 – 2ad \implies 2ad = u^2 \frac{(n – 1)^2}{n^2}$.
Dividing the Phase 2 equation by the Phase 1 equation:
$\frac{2ad}{2ax} = \frac{u^2 \frac{(n – 1)^2}{n^2}}{u^2 \frac{2n – 1}{n^2}} \implies \frac{d}{x} = \frac{(n – 1)^2}{2n – 1}$.
$d = x \frac{(n – 1)^2}{2n – 1}$.
Special Case: If the bullet loses half its speed ($n = 2$) in distance $x$, the additional distance is $d = x \frac{(2 – 1)^2}{2(2) – 1} = \frac{x}{3}$.
Example 3 (Two-Vehicle Relative Motion & Collision Avoidance):
Car $A$ is moving at $u_A = 30\text{ m/s}$ along a straight highway, trailing Car $B$ which is moving at $u_B = 10\text{ m/s}$ in the same direction. When the separation between them is $d = 80\text{ m}$, the driver of Car $A$ applies brakes, producing a constant deceleration $a_A = 2.0\text{ m/s}^2$. Car $B$ continues at constant speed.
(a) Determine whether a collision occurs.
(b) If a collision occurs, find the time of impact. If not, find the minimum separation.
Solution:
Analyze the motion in the reference frame of Car $B$:
Initial relative velocity: $u_{\text{rel}} = u_A – u_B = 30 – 10 = 20\text{ m/s}$.
Relative acceleration: $a_{\text{rel}} = a_A – a_B = -2.0 – 0 = -2.0\text{ m/s}^2$.
Relative stopping distance (to reduce relative speed to zero):
$v_{\text{rel}}^2 = u_{\text{rel}}^2 + 2 a_{\text{rel}} s_{\text{rel}} \implies 0 = 20^2 – 2(2.0) s_{\text{rel}} \implies 4 s_{\text{rel}} = 400 \implies s_{\text{rel}} = 100\text{ m}$.
(a) Since $s_{\text{rel}} = 100\text{ m} > d = 80\text{ m}$, Car $A$ cannot stop relative to Car $B$ before covering the gap. A collision occurs!
(b) Relative displacement at impact is $s_{\text{rel}} = 80\text{ m}$:
$s_{\text{rel}} = u_{\text{rel}} t + \frac{1}{2}a_{\text{rel}} t^2 \implies 80 = 20t – \frac{1}{2}(2.0)t^2 \implies t^2 – 20t + 80 = 0$.
Using the quadratic formula:
$t = \frac{20 \pm \sqrt{400 – 320}}{2} = \frac{20 \pm \sqrt{80}}{2} = 10 \pm 2\sqrt{5}\text{ s}$.
The collision occurs at the smaller root:
$t = 10 – 2\sqrt{5} \approx 10 – 4.47 = 5.53\text{ seconds}$.
Example 4 (Calculus with Spatial Acceleration $a(x) = v \frac{dv}{dx}$):
A particle starts from rest at $x = 0$ and moves along the $x$-axis subject to an acceleration $a(x) = 3x^2 + 2x\text{ m/s}^2$. Find:
(a) The velocity $v(x)$ as an explicit function of position $x$.
(b) The speed of the particle when it reaches $x = 2\text{ m}$.
Solution:
(a) Using $a = v \frac{dv}{dx}$:
$v \frac{dv}{dx} = 3x^2 + 2x \implies v\,dv = (3x^2 + 2x)\,dx$.
Integrating from initial state ($x = 0, v = 0$) to position $x$:
$\int_{0}^{v} v\,dv = \int_{0}^{x} (3x^2 + 2x)\,dx \implies \left[\frac{v^2}{2}\right]_{0}^{v} = \left[x^3 + x^2\right]_{0}^{x}$
$\frac{v^2}{2} = x^3 + x^2 \implies v^2 = 2x^2(x + 1)$.
$v(x) = x \sqrt{2(x + 1)}$.
(b) At $x = 2\text{ m}$:
$v(2) = 2 \sqrt{2(2 + 1)} = 2 \sqrt{6}\text{ m/s} \approx 4.90\text{ m/s}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A bullet loses $\frac{1}{3}$ of its velocity after penetrating $4\text{ cm}$ into a wooden plank. Assuming constant deceleration, how much further will it penetrate before coming to rest?
(A) $3.2\text{ cm}$
(B) $3.6\text{ cm}$
(C) $4.0\text{ cm}$
(D) $4.8\text{ cm}$
Problem 2 (JEE Main – Single Correct):
A car traveling at speed $u$ can be brought to rest in a minimum distance $d$ by applying brakes. If the same car travels at speed $3u$, the minimum stopping distance under identical braking deceleration is:
(A) $3d$
(B) $6d$
(C) $9d$
(D) $12d$
Problem 3 (JEE Main – Single Correct):
A particle moving with uniform acceleration along a straight line has velocities $10\text{ m/s}$ and $20\text{ m/s}$ at two points $A$ and $B$. The velocity of the particle at the midpoint between $A$ and $B$ is:
(A) $15\text{ m/s}$
(B) $5\sqrt{10}\text{ m/s}$
(C) $10\sqrt{2}\text{ m/s}$
(D) $16\text{ m/s}$
Problem 4 (JEE Main – Single Correct):
A particle starts from rest and moves with uniform acceleration $a$ over distance $s_1$ and attains speed $v$. It then covers distance $s_2$ with acceleration $2a$ and attains speed $2v$. The ratio $\frac{s_2}{s_1}$ is:
(A) $1 : 1$
(B) $3 : 2$
(C) $2 : 1$
(D) $3 : 1$
Problem 5 (JEE Main – Single Correct):
A body moves with constant acceleration along a straight line. If it travels a distance $s$ in time $t$ while its speed increases from $u$ to $v$, then the distance $s$ is given by:
(A) $s = (v – u) t$
(B) $s = \left(\frac{u + v}{2}\right) t$
(C) $s = \sqrt{u v} \cdot t$
(D) $s = \left(\frac{v^2 + u^2}{2}\right) t$
Problem 6 (JEE Advanced – One or More Correct):
A particle moves along a straight line with uniform acceleration $a$. If initial speed is $u$ and final speed is $v$ after displacement $s$, which of the following statements is/are correct?
(A) $v^2 = u^2 + 2as$ is valid strictly when acceleration is constant.
(B) The velocity at the spatial midpoint of displacement is $\sqrt{\frac{u^2 + v^2}{2}}$.
(C) The velocity at the temporal midpoint (half-time) of the motion is $\frac{u + v}{2}$.
(D) The average velocity over displacement $s$ is $\frac{u + v}{2}$.
Problem 7 (JEE Advanced – One or More Correct):
A body starts from rest at $x = 0$ and moves with uniform acceleration along the $x$-axis. Which of the following statements is/are TRUE?
(A) The ratio of time taken to cover the first half of distance to the second half is $\frac{1}{\sqrt{2} – 1}$.
(B) The ratio of distance covered in the first half of total time to the second half is $1 : 3$.
(C) The speed acquired after half the total time is half the final speed.
(D) The speed acquired after half the total distance is $\frac{1}{\sqrt{2}}$ of the final speed.
Problem 8 (JEE Advanced – One or More Correct):
A vehicle moving with initial velocity $v_0$ experiences a non-linear resistive deceleration $a = -\alpha v^2$, where $\alpha$ is a positive constant. Which of the following statements is/are correct?
(A) Velocity as a function of position $x$ is $v(x) = v_0 e^{-\alpha x}$.
(B) The vehicle comes to rest only as $x \to \infty$.
(C) The vehicle comes to rest in a finite distance.
(D) The magnitude of deceleration decreases as velocity decreases.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A train passes an electric pole with uniform acceleration. The front locomotive passes at $12\text{ m/s}$ and the rear end passes at $16\text{ m/s}$. The speed with which the middle of the train passes the pole is $V\text{ m/s}$. Find the integer value of $V^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A bullet loses $50\%$ of its velocity on penetrating $6.0\text{ cm}$ into a wooden target. The additional distance it penetrates in $\text{cm}$ before coming to rest is $d$. Find the numerical value of $10 \times d$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (C) | 3. (B) | 4. (B) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, D) | 9. 200 | 10. 20
Solution 1:
Formula: $d = x \frac{(n – 1)^2}{2n – 1}$.
Here $n = 3, x = 4\text{ cm}$:
$d = 4 \times \frac{(3 – 1)^2}{2(3) – 1} = 4 \times \frac{4}{5} = \frac{16}{5} = 3.2\text{ cm}$.
Correct Answer: (A)
Solution 2:
$d_{\text{stop}} = \frac{u^2}{2a}$.
For initial speed $3u$: $d’ = \frac{(3u)^2}{2a} = 9 \left(\frac{u^2}{2a}\right) = 9d$.
Correct Answer: (C)
Solution 3:
$v_{\text{mid}} = \sqrt{\frac{u^2 + v^2}{2}} = \sqrt{\frac{10^2 + 20^2}{2}} = \sqrt{\frac{100 + 400}{2}} = \sqrt{250} = 5\sqrt{10}\text{ m/s}$.
Correct Answer: (B)
Solution 4:
– Stage 1: $v^2 = 0 + 2as_1 \implies s_1 = \frac{v^2}{2a}$.
– Stage 2: $(2v)^2 = v^2 + 2(2a)s_2 \implies 4v^2 – v^2 = 4as_2 \implies 3v^2 = 4as_2 \implies s_2 = \frac{3v^2}{4a}$.
Ratio $\frac{s_2}{s_1} = \frac{3v^2 / (4a)}{v^2 / (2a)} = \frac{3}{2}$.
Correct Answer: (B)
Solution 5:
For uniform acceleration, $v_{\text{avg}} = \frac{u + v}{2}$, so $s = v_{\text{avg}} t = \left(\frac{u + v}{2}\right) t$.
Correct Answer: (B)
Solution 6:
All four statements (A, B, C, D) are fundamental principles of uniformly accelerated rectilinear motion.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) $\frac{s}{2} = \frac{1}{2}a t_1^2 \implies t_1 = \sqrt{\frac{s}{a}}$. Total time $T = \sqrt{\frac{2s}{a}}$. Second half time $t_2 = T – t_1 = \sqrt{\frac{s}{a}}(\sqrt{2} – 1)$. Ratio $\frac{t_1}{t_2} = \frac{1}{\sqrt{2} – 1}$ (True).
– (B) Distance in first half of time: $s_1 = \frac{1}{2}a(T/2)^2 = \frac{s}{4}$. Distance in second half: $s_2 = s – \frac{s}{4} = \frac{3s}{4}$. Ratio $= 1 : 3$ (True).
– (C) $v(T/2) = a(T/2) = \frac{1}{2}(aT) = \frac{v_{\text{final}}}{2}$ (True).
– (D) $v_{\text{mid}} = \sqrt{\frac{0 + v_{\text{final}}^2}{2}} = \frac{v_{\text{final}}}{\sqrt{2}}$ (True).
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 8:
$a = v \frac{dv}{dx} = -\alpha v^2 \implies \frac{dv}{v} = -\alpha\,dx$.
Integrating: $\ln\left(\frac{v}{v_0}\right) = -\alpha x \implies v(x) = v_0 e^{-\alpha x}$.
– As $x \to \infty$, $v \to 0$ (A and B are True; C is False).
– Deceleration magnitude is $\alpha v^2$, which decreases as $v$ decreases (D is True).
Correct Answer: (A, B, D)
Solution 9:
$V^2 = \frac{u^2 + v^2}{2} = \frac{12^2 + 16^2}{2} = \frac{144 + 256}{2} = \frac{400}{2} = 200$.
Correct Answer: 200
Solution 10:
$n = 2, x = 6.0\text{ cm}$.
$d = x \frac{(n – 1)^2}{2n – 1} = 6.0 \times \frac{(2 – 1)^2}{2(2) – 1} = \frac{6.0}{3} = 2.0\text{ cm}$.
$10 \times d = 10 \times 2.0 = 20$.
Correct Answer: 20