Concept Card: Acceleration & Uniformly Accelerated Motion
1. Definition & Mathematical Properties of Acceleration:
Acceleration ($\vec{a}$) is defined as the time rate of change of velocity:
$\vec{a} = \frac{d\vec{v}}{dt}$
It is a vector quantity with dimensions $[M^0 L^1 T^{-2}]$ and SI unit $\text{m/s}^2$.
- Average Acceleration ($\vec{a}_{\text{avg}}$):
$\vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_f – \vec{v}_i}{t_f – t_i}$ - Instantaneous Acceleration ($\vec{a}$):
$\vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}$ - Spatial Chain Rule Form (1D Rectilinear Motion):
$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$
(Crucial for problems where acceleration is given as a function of position $x$, rather than time $t$).
2. Uniformly Accelerated Motion ($a = \text{constant}$):
- Derivation of the First Kinematic Equation ($v = u + at$):
By definition: $a = \frac{dv}{dt} \implies dv = a\,dt$.
Integrating with initial condition $v(0) = u$ at $t = 0$ to $v(t) = v$ at time $t$:
$\int_{u}^{v} dv = \int_{0}^{t} a\,dt \implies [v]_{u}^{v} = a [t]_{0}^{t} \implies v – u = at \implies v = u + at$. - Derivation of the Second Kinematic Equation ($s = ut + \frac{1}{2}at^2$):
Instantaneous velocity is rate of displacement: $v = \frac{ds}{dt} \implies ds = v\,dt$.
Substituting $v = u + at$:
$ds = (u + at)\,dt$.
Integrating from $s = 0$ at $t = 0$ to $s(t) = s$ at time $t$:
$\int_{0}^{s} ds = \int_{0}^{t} (u + at)\,dt \implies s = \left[ut + \frac{1}{2}at^2\right]_{0}^{t} \implies s = ut + \frac{1}{2}at^2$. - Displacement in the $n$-th Second ($s_n$):
The displacement occurring during the specific one-second interval between $t = (n – 1)$ and $t = n$:
$s_n = s(n) – s(n – 1) = \left[un + \frac{1}{2}an^2\right] – \left[u(n – 1) + \frac{1}{2}a(n – 1)^2\right]$
$s_n = u + \frac{a}{2}(2n – 1)$.
3. Critical Traps & Insights for JEE:
- Deceleration / Retardation: Occurs if and only if velocity and acceleration have opposite signs ($\vec{a} \cdot \vec{v} < 0$). Negative acceleration alone does not mean slowing down (e.g., if $v = -10\text{ m/s}$ and $a = -3\text{ m/s}^2$, the body is accelerating in the negative direction).
- Zero Velocity with Non-Zero Acceleration: A particle can momentarily come to rest ($v = 0$) while possessing finite acceleration (e.g., at the crest of vertical projectile motion, $v = 0$ while $a = -g$).
- Boundary of Validity: The formulas $v = u + at$ and $s = ut + \frac{1}{2}at^2$ hold only when acceleration is strictly constant. If acceleration varies with time ($a(t)$) or position ($a(x)$), calculus integration must be used.
Solved Examples
Example 1 (Basic Kinematic Equations & $n$-th Second Displacement):
A racing car starts from rest and moves along a straight track with constant acceleration $a = 3.0\text{ m/s}^2$.
Find:
(a) Its velocity after $5\text{ seconds}$.
(b) The total distance traversed in the first $5\text{ seconds}$.
(c) The distance covered specifically in the $5$-th second of motion.
Solution:
Given: $u = 0$, $a = 3.0\text{ m/s}^2$.
(a) $v = u + at = 0 + 3.0(5) = 15.0\text{ m/s}$.
(b) $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3.0)(5^2) = 1.5 \times 25 = 37.5\text{ m}$.
(c) Distance in the $5$-th second ($n = 5$):
$s_n = u + \frac{a}{2}(2n – 1) = 0 + \frac{3.0}{2}(2 \times 5 – 1) = 1.5 \times 9 = 13.5\text{ m}$.
Verification: $s(5) – s(4) = 37.5 – \left[\frac{1}{2} \times 3.0 \times 16\right] = 37.5 – 24.0 = 13.5\text{ m}$.
Example 2 (Two-Phase Motion: Acceleration Followed by Deceleration):
A commuter train starts from rest at station $A$, accelerates uniformly at $\alpha = 2.0\text{ m/s}^2$ for time $t_1$ to reach top speed $v_{\text{max}}$, and immediately decelerates uniformly at $\beta = 1.0\text{ m/s}^2$ for time $t_2$ to come to rest at station $B$. The track distance between stations is $S = 1200\text{ m}$. Determine:
(a) The total transit time $T = t_1 + t_2$.
(b) The maximum velocity $v_{\text{max}}$.
Solution:
1. Express time intervals in terms of $v_{\text{max}}$:
During acceleration: $v_{\text{max}} = \alpha t_1 \implies t_1 = \frac{v_{\text{max}}}{\alpha}$.
During deceleration: $0 = v_{\text{max}} – \beta t_2 \implies t_2 = \frac{v_{\text{max}}}{\beta}$.
Total time: $T = t_1 + t_2 = v_{\text{max}}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = v_{\text{max}} \frac{\alpha + \beta}{\alpha \beta}$.
2. Total distance is the area under the triangular $v-t$ graph:
$S = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} T v_{\text{max}} = \frac{1}{2} \left[v_{\text{max}} \frac{\alpha + \beta}{\alpha \beta}\right] v_{\text{max}} = \frac{v_{\text{max}}^2 (\alpha + \beta)}{2\alpha\beta}$.
3. Solve for $v_{\text{max}}$:
$v_{\text{max}} = \sqrt{\frac{2 S \alpha \beta}{\alpha + \beta}} = \sqrt{\frac{2 \times 1200 \times 2.0 \times 1.0}{2.0 + 1.0}} = \sqrt{\frac{4800}{3}} = \sqrt{1600} = 40\text{ m/s}$.
4. Calculate total time $T$:
$T = \frac{2S}{v_{\text{max}}} = \frac{2 \times 1200}{40} = 60\text{ seconds}$.
Example 3 (Reaction Time & Stopping Distance):
A driver is traveling along a straight highway at $72\text{ km/h}$ ($20\text{ m/s}$). Spotting an obstacle ahead, the driver reacts after a mental reaction time $t_r = 0.4\text{ s}$ and slams on the brakes, producing a constant deceleration $a = 5.0\text{ m/s}^2$.
(a) Find the total stopping distance of the vehicle.
(b) If the obstacle was originally $45\text{ m}$ ahead, does a collision occur? If yes, find the impact speed.
Solution:
(a) During the reaction time $t_r = 0.4\text{ s}$, the vehicle continues with constant velocity $u = 20\text{ m/s}$:
$d_r = u \cdot t_r = 20 \times 0.4 = 8.0\text{ m}$.
During braking, initial speed is $u = 20\text{ m/s}$, final speed $v = 0$, deceleration $a = -5.0\text{ m/s}^2$:
$v^2 = u^2 + 2as_b \implies 0 = 20^2 + 2(-5.0)d_b \implies 10 d_b = 400 \implies d_b = 40.0\text{ m}$.
Total stopping distance: $d_{\text{stop}} = d_r + d_b = 8.0 + 40.0 = 48.0\text{ m}$.
(b) Since the obstacle is at $45.0\text{ m} < 48.0\text{ m}$, the car strikes the obstacle.
Distance available for braking: $s_b = 45.0 – 8.0 = 37.0\text{ m}$.
Speed at impact:
$v^2 = u^2 + 2as_b = 20^2 + 2(-5.0)(37.0) = 400 – 370 = 30\text{ m}^2/\text{s}^2$.
$v = \sqrt{30}\text{ m/s} \approx 5.48\text{ m/s}$.
Example 4 (Calculus with Spatial Acceleration $v \frac{dv}{dx}$):
A body moves along the $x$-axis subject to a resistive deceleration given by $a = -k v$, where $k = 0.5\text{ s}^{-1}$ and $v$ is instantaneous speed. If the initial speed at $x = 0$ is $u = 10\text{ m/s}$, calculate the total distance the body travels before coming completely to rest.
Solution:
Express acceleration using the spatial derivative form $a = v \frac{dv}{dx}$:
$v \frac{dv}{dx} = -k v$.
For non-zero velocity ($v \neq 0$), divide by $v$:
$\frac{dv}{dx} = -k \implies dv = -k\,dx$.
Integrate from initial state ($x = 0, v = u$) to final state ($x = x_{\text{max}}, v = 0$):
$\int_{u}^{0} dv = -k \int_{0}^{x_{\text{max}}} dx \implies [v]_{u}^{0} = -k [x]_{0}^{x_{\text{max}}} \implies 0 – u = -k x_{\text{max}}$.
$x_{\text{max}} = \frac{u}{k}$.
Substituting $u = 10\text{ m/s}$ and $k = 0.5\text{ s}^{-1}$:
$x_{\text{max}} = \frac{10}{0.5} = 20\text{ m}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A body starts from rest and moves with uniform acceleration. The ratio of distances covered by the body in the $1$-st, $2$-nd, and $3$-rd seconds of its motion is:
(A) $1 : 2 : 3$
(B) $1 : 4 : 9$
(C) $1 : 3 : 5$
(D) $1 : 5 : 9$
Problem 2 (JEE Main – Single Correct):
A particle moving with uniform acceleration covers $12\text{ m}$ in the $2$-nd second and $20\text{ m}$ in the $4$-th second of its motion. The initial velocity of the particle is:
(A) $2\text{ m/s}$
(B) $4\text{ m/s}$
(C) $6\text{ m/s}$
(D) $8\text{ m/s}$
Problem 3 (JEE Main – Single Correct):
A car starts from rest and accelerates with a constant acceleration of $2\text{ m/s}^2$. The distance traveled by the car specifically during the $4$-th second is:
(A) $7\text{ m}$
(B) $8\text{ m}$
(C) $9\text{ m}$
(D) $16\text{ m}$
Problem 4 (JEE Main – Single Correct):
A train starts from rest with acceleration $\alpha$ for time $t_1$, moves with constant velocity for time $t_2$, and finally decelerates to rest at rate $\beta$ in time $t_3$. If the maximum velocity attained is $v_{\text{max}}$, the total distance covered is:
(A) $\frac{1}{2} v_{\text{max}} (t_1 + t_2 + t_3)$
(B) $v_{\text{max}} \left[\frac{t_1 + t_3}{2} + t_2\right]$
(C) $v_{\text{max}} (t_1 + t_2 + t_3)$
(D) $\frac{1}{2} v_{\text{max}} (t_1 + t_3)$
Problem 5 (JEE Main – Single Correct):
A body moving with uniform acceleration covers a distance of $20\text{ m}$ in the first $2\text{ seconds}$ and $160\text{ m}$ in the next $4\text{ seconds}$. The initial velocity of the body is:
(A) $0\text{ m/s}$
(B) $5\text{ m/s}$
(C) $10\text{ m/s}$
(D) $2.5\text{ m/s}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct for rectilinear motion with uniform acceleration $a$?
(A) The average velocity over any time interval equals the arithmetic mean of the initial and final velocities: $v_{\text{avg}} = \frac{u + v}{2}$.
(B) The position-time graph is a parabola opening upwards if $a > 0$.
(C) The slope of the velocity-time graph represents the acceleration at that instant.
(D) If velocity and acceleration have opposite signs, the speed of the particle decreases.
Problem 7 (JEE Advanced – One or More Correct):
A particle moves along the $x$-axis with constant acceleration $a$. If it covers distances $s_1$ in time $t$ and $s_2$ in the next equal time interval $t$, which of the following relations is/are correct?
(A) The acceleration is given by $a = \frac{s_2 – s_1}{t^2}$.
(B) The velocity at the end of the first interval is $v_1 = \frac{s_1 + s_2}{2t}$.
(C) The average velocity over the entire $2t$ interval is $v_{\text{avg}} = \frac{s_1 + s_2}{2t}$.
(D) The initial velocity is $u = \frac{3s_1 – s_2}{2t}$.
Problem 8 (JEE Advanced – One or More Correct):
For a particle starting from rest and moving with constant acceleration $a$ along a straight line:
(A) The ratio of distance covered in the $n$-th second to the total distance covered in $n$ seconds is $\frac{2n – 1}{n^2}$.
(B) As $n \to \infty$, the fraction $\frac{2n – 1}{n^2}$ approaches $0$.
(C) The displacement in the $1$-st second is equal to half the acceleration ($\frac{a}{2}$).
(D) The total displacement in the first $n$ seconds is proportional to $n^2$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A particle starts from rest with constant acceleration $a = 4.0\text{ m/s}^2$. The displacement covered by the particle in the $3$-rd second is $S_3\text{ m}$. Find the numerical value of $S_3$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A car moving with constant acceleration covers $30\text{ m}$ in $2\text{ seconds}$ and $80\text{ m}$ in the next $2\text{ seconds}$. The acceleration of the car in $\text{m/s}^2$ is $a$. Determine the numerical value of $10 \times a$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (A) | 4. (B) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 10 | 10. 125
Solution 1:
According to Galileo’s Law of Odd Numbers: For a body starting from rest ($u = 0$) with uniform acceleration, the displacement in the $n$-th second is $s_n = \frac{a}{2}(2n – 1)$.
For $n = 1, 2, 3$:
$s_1 : s_2 : s_3 = (2 \times 1 – 1) : (2 \times 2 – 1) : (2 \times 3 – 1) = 1 : 3 : 5$.
Correct Answer: (C)
Solution 2:
Formula: $s_n = u + \frac{a}{2}(2n – 1)$.
For $n = 2$: $12 = u + \frac{a}{2}(3) \implies 12 = u + 1.5a$.
For $n = 4$: $20 = u + \frac{a}{2}(7) \implies 20 = u + 3.5a$.
Subtracting the equations: $20 – 12 = (3.5a – 1.5a) \implies 8 = 2a \implies a = 4\text{ m/s}^2$.
Substitute $a = 4$ into the first equation:
$12 = u + 1.5(4) \implies 12 = u + 6 \implies u = 6\text{ m/s}$.
Correct Answer: (C)
Solution 3:
$s_n = u + \frac{a}{2}(2n – 1)$.
For $u = 0, a = 2\text{ m/s}^2, n = 4$:
$s_4 = 0 + \frac{2}{2}(2 \times 4 – 1) = 1 \times 7 = 7\text{ m}$.
Correct Answer: (A)
Solution 4:
The $v-t$ curve forms a trapezoid:
– Lower base: $T = t_1 + t_2 + t_3$
– Upper base: $t_2$
– Height: $v_{\text{max}}$
$\text{Area} = \frac{1}{2}(\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} (t_1 + 2t_2 + t_3) v_{\text{max}} = v_{\text{max}} \left[\frac{t_1 + t_3}{2} + t_2\right]$.
Correct Answer: (B)
Solution 5:
– In first $2\text{ s}$: $s_1 = 20\text{ m} \implies 20 = u(2) + \frac{1}{2}a(2^2) \implies 20 = 2u + 2a \implies u + a = 10$.
– In total $6\text{ s}$: total distance $s = 20 + 160 = 180\text{ m} \implies 180 = u(6) + \frac{1}{2}a(6^2) \implies 180 = 6u + 18a \implies u + 3a = 30$.
Subtracting equations: $2a = 20 \implies a = 10\text{ m/s}^2$.
Substitute into $u + a = 10 \implies u + 10 = 10 \implies u = 0\text{ m/s}$.
Correct Answer: (A)
Solution 6:
– (A) True: $v_{\text{avg}} = \frac{s}{t} = \frac{ut + \frac{1}{2}at^2}{t} = u + \frac{1}{2}at = \frac{u + (u + at)}{2} = \frac{u + v}{2}$.
– (B) True: $x(t) = x_0 + ut + \frac{1}{2}at^2$ is parabolic opening upwards when $a > 0$.
– (C) True: By definition, $\frac{dv}{dt} = \text{slope} = a$.
– (D) True: When $\vec{a} \cdot \vec{v} < 0$, speed $|\vec{v}|$ decreases.
Correct Answer: (A, B, C, D)
Solution 7:
– $s_1 = ut + \frac{1}{2}at^2$
– $s_2 = (u + at)t + \frac{1}{2}at^2 = ut + \frac{3}{2}at^2$
Subtracting gives $s_2 – s_1 = at^2 \implies a = \frac{s_2 – s_1}{t^2}$ (A is True).
Velocity at end of first interval: $v_1 = u + at = \frac{s_1}{t} – \frac{1}{2}at + at = \frac{s_1}{t} + \frac{1}{2}at = \frac{s_1}{t} + \frac{s_2 – s_1}{2t} = \frac{s_1 + s_2}{2t}$ (B is True).
Average velocity over $2t$: $v_{\text{avg}} = \frac{s_1 + s_2}{2t}$ (C is True).
Initial velocity: $u = \frac{s_1}{t} – \frac{1}{2}at = \frac{s_1}{t} – \frac{s_2 – s_1}{2t} = \frac{3s_1 – s_2}{2t}$ (D is True).
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) $\frac{s_n}{s_{\text{total}}} = \frac{\frac{a}{2}(2n – 1)}{\frac{1}{2}an^2} = \frac{2n – 1}{n^2}$ (True).
– (B) $\lim_{n \to \infty} \frac{2n – 1}{n^2} = \lim_{n \to \infty} \left(\frac{2}{n} – \frac{1}{n^2}\right) = 0$ (True).
– (C) For $n = 1$: $s_1 = \frac{a}{2}(2 \times 1 – 1) = \frac{a}{2}$ (True).
– (D) $s = \frac{1}{2}an^2 \propto n^2$ (True).
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 9:
$S_3 = u + \frac{a}{2}(2 \times 3 – 1) = 0 + \frac{4.0}{2}(5) = 2.0 \times 5 = 10\text{ m}$.
Correct Answer: 10
Solution 10:
– First $2\text{ s}$: $30 = 2u + 2a \implies u + a = 15$.
– Total $4\text{ s}$: $30 + 80 = 110\text{ m} \implies 110 = 4u + 8a \implies u + 2a = 27.5$.
Subtracting: $a = 27.5 – 15 = 12.5\text{ m/s}^2$.
$10 \times a = 10 \times 12.5 = 125$.
Correct Answer: 125