Concept Card: Speed, Velocity & Types of Motion
1. Speed vs. Velocity:
- Speed ($v$): The rate at which distance is covered with respect to time. It is a scalar quantity ($v \ge 0$) with dimensions $[L T^{-1}]$ and SI unit $\text{m/s}$.
- Velocity ($\vec{v}$): The rate of change of the position vector with respect to time ($\vec{v} = \frac{d\vec{r}}{dt}$). It is a vector quantity possessing both magnitude and direction. In 1D rectilinear motion, direction is indicated by an algebraic sign ($+$ or $-$).
2. Average Quantities vs. Instantaneous Quantities:
- Average Speed ($v_{\text{avg}}$):
$v_{\text{avg}} = \frac{\text{Total Distance Traversed}}{\text{Total Time Elapsed}} = \frac{\Delta s}{\Delta t}$- Case 1 – Harmonic Mean (Equal Distances): If an object covers distance $d_1$ at speed $v_1$ and $d_2$ at speed $v_2$, then:
$v_{\text{avg}} = \frac{d_1 + d_2}{\frac{d_1}{v_1} + \frac{d_2}{v_2}} \implies \text{For equal distances } (d_1 = d_2 = d): \quad v_{\text{avg}} = \frac{2 v_1 v_2}{v_1 + v_2}$ - Case 2 – Arithmetic Mean (Equal Time Intervals): If an object travels with speed $v_1$ for time $t_1$ and $v_2$ for time $t_2$, then:
$v_{\text{avg}} = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2} \implies \text{For equal times } (t_1 = t_2 = t): \quad v_{\text{avg}} = \frac{v_1 + v_2}{2}$
- Case 1 – Harmonic Mean (Equal Distances): If an object covers distance $d_1$ at speed $v_1$ and $d_2$ at speed $v_2$, then:
- Average Velocity ($\vec{v}_{\text{avg}}$):
$\vec{v}_{\text{avg}} = \frac{\text{Net Displacement}}{\text{Total Time Elapsed}} = \frac{\vec{r}_f – \vec{r}_i}{\Delta t} = \frac{\Delta \vec{r}}{\Delta t}$
Fundamental Inequality: $v_{\text{avg}} \ge |\vec{v}_{\text{avg}}|$. - Instantaneous Velocity ($\vec{v}$):
$\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}$
In 1D along the $x$-axis: $v(t) = \frac{dx}{dt}$. Geometrically, instantaneous velocity represents the slope of the tangent to the position-time ($x-t$) curve. - Instantaneous Speed:
Instantaneous speed is strictly defined as the magnitude of instantaneous velocity:
$v_{\text{inst}} = |\vec{v}| = \left|\frac{dx}{dt}\right|$.
3. Uniform vs. Non-Uniform Motion:
- Uniform Motion: A particle moves with constant velocity (constant speed along a fixed unchanging straight line). Acceleration $\vec{a} = 0$. Position varies linearly with time: $x(t) = x_0 + v t$. Average velocity over any interval equals instantaneous velocity at every point.
- Non-Uniform Motion: Velocity changes with time (in magnitude, direction, or both). Acceleration $\vec{a} \neq 0$.
Solved Examples
Example 1 (Multi-Segment Average Speed – Harmonic Mean):
A car travels from point $A$ to point $B$. It traverses the first one-third of the total distance at $v_1 = 20\text{ km/h}$, the second one-third at $v_2 = 30\text{ km/h}$, and the final one-third at $v_3 = 60\text{ km/h}$. Find the average speed for the entire journey.
Solution:
Let total distance be $3d$. Each segment has length $d$.
Time for segment 1: $t_1 = \frac{d}{20}$.
Time for segment 2: $t_2 = \frac{d}{30}$.
Time for segment 3: $t_3 = \frac{d}{60}$.
Total time: $T = d\left(\frac{1}{20} + \frac{1}{30} + \frac{1}{60}\right) = d\left(\frac{3}{60} + \frac{2}{60} + \frac{1}{60}\right) = d\left(\frac{6}{60}\right) = \frac{d}{10}$.
Average speed: $v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{3d}{\frac{d}{10}} = 30\text{ km/h}$.
Example 2 (Calculus-Based Velocity, Turning Points, and Average Quantities):
The position of a particle along the $x$-axis is given by $x(t) = 2t^3 – 15t^2 + 24t + 10\text{ m}$.
(a) Find the instantaneous velocity at $t = 2\text{ s}$.
(b) Find the timestamps when the particle momentarily comes to rest.
(c) Find average velocity between $t = 0$ and $t = 5\text{ s}$.
(d) Find average speed between $t = 0$ and $t = 5\text{ s}$.
Solution:
(a) $v(t) = \frac{dx}{dt} = 6t^2 – 30t + 24\text{ m/s}$.
At $t = 2\text{ s}$: $v(2) = 6(4) – 30(2) + 24 = 24 – 60 + 24 = -12\text{ m/s}$.
(b) $v(t) = 0 \implies 6(t^2 – 5t + 4) = 0 \implies (t – 1)(t – 4) = 0$.
The particle momentarily comes to rest at $t = 1\text{ s}$ and $t = 4\text{ s}$.
(c) Evaluate positions:
$x(0) = 10\text{ m}$
$x(1) = 2(1) – 15(1) + 24(1) + 10 = 21\text{ m}$
$x(4) = 2(64) – 15(16) + 24(4) + 10 = 128 – 240 + 96 + 10 = -6\text{ m}$
$x(5) = 2(125) – 15(25) + 24(5) + 10 = 250 – 375 + 120 + 10 = 5\text{ m}$.
Displacement from $t = 0$ to $t = 5\text{ s}$: $\Delta x = x(5) – x(0) = 5 – 10 = -5\text{ m}$.
Average velocity: $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{-5\text{ m}}{5\text{ s}} = -1.0\text{ m/s}$.
(d) Total Distance:
$s_1 (0 \to 1\text{ s}) = |21 – 10| = 11\text{ m}$
$s_2 (1 \to 4\text{ s}) = |-6 – 21| = 27\text{ m}$
$s_3 (4 \to 5\text{ s}) = |5 – (-6)| = 11\text{ m}$
Total distance: $s = 11 + 27 + 11 = 49\text{ m}$.
Average speed: $\frac{49\text{ m}}{5\text{ s}} = 9.8\text{ m/s}$.
Example 3 (2D Motion in a Plane):
A particle moves in the $x-y$ plane with $x(t) = 4t^2$ and $y(t) = 3t^2 + 5$.
(a) Find velocity vector $\vec{v}(t) = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = 8t\hat{i} + 6t\hat{j}$.
At $t = 2\text{ s}$: $\vec{v}(2) = 16\hat{i} + 12\hat{j}\text{ m/s}$.
Magnitude: $v = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20\text{ m/s}$.
Direction: $\tan\theta = \frac{v_y}{v_x} = \frac{12}{16} = \frac{3}{4} \implies \theta \approx 37^\circ$ to $+x$ axis.
(b) From $t = 0$ to $t = 2\text{ s}$:
$\vec{r}_i = 0\hat{i} + 5\hat{j}\text{ m}$; $\vec{r}_f = 16\hat{i} + 17\hat{j}\text{ m}$.
$\Delta \vec{r} = 16\hat{i} + 12\hat{j}\text{ m}$.
Average velocity: $\vec{v}_{\text{avg}} = \frac{16\hat{i} + 12\hat{j}}{2} = 8\hat{i} + 6\hat{j}\text{ m/s}$ (Magnitude $= 10\text{ m/s}$).
Since the trajectory is a straight line $y – 5 = \frac{3}{4}x$ without reversal, distance $= |\Delta \vec{r}| = 20\text{ m}$.
Average speed $= \frac{20\text{ m}}{2\text{ s}} = 10\text{ m/s}$.
Example 4 (Position-Dependent Velocity):
A particle moves in a straight line with velocity $v(x) = \alpha \sqrt{x}$. If $x(0) = 0$ at $t = 0$:
(a) $v = \frac{dx}{dt} = \alpha \sqrt{x} \implies \frac{dx}{\sqrt{x}} = \alpha dt$.
Integrating from $0$ to $x$ and $0$ to $t$: $2\sqrt{x} = \alpha t \implies x(t) = \frac{\alpha^2}{4} t^2$.
$v(t) = \frac{dx}{dt} = \frac{\alpha^2}{2} t$.
(b) Time to travel distance $S$: $2\sqrt{S} = \alpha t_S \implies t_S = \frac{2\sqrt{S}}{\alpha}$.
Average velocity over distance $S$: $v_{\text{avg}} = \frac{S}{t_S} = \frac{S}{\frac{2\sqrt{S}}{\alpha}} = \frac{\alpha}{2}\sqrt{S}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A vehicle travels half the total distance with speed $v_1$ and the remaining half with speed $v_2$. The average speed of the vehicle is:
(A) $\frac{v_1 + v_2}{2}$
(B) $\frac{2 v_1 v_2}{v_1 + v_2}$
(C) $\sqrt{v_1 v_2}$
(D) $\frac{v_1 v_2}{v_1 + v_2}$
Problem 2 (JEE Main – Single Correct):
The displacement of a particle moving along the $x$-axis is given by $x = a t^2 – b t^3$. The time at which the instantaneous velocity of the particle becomes zero is:
(A) $\frac{a}{b}$
(B) $\frac{2a}{3b}$
(C) $\frac{a}{3b}$
(D) $\frac{3a}{2b}$
Problem 3 (JEE Main – Single Correct):
A particle moves around a circle of radius $R$ with constant speed $v$. The magnitude of average velocity of the particle as it completes half a revolution is:
(A) $v$
(B) $\frac{2v}{\pi}$
(C) $\frac{\pi v}{2}$
(D) $0$
Problem 4 (JEE Main – Single Correct):
A body moves for the first half of its total transit time with speed $40\text{ km/h}$ and for the second half of the time with speed $60\text{ km/h}$. The average speed of the body is:
(A) $48\text{ km/h}$
(B) $50\text{ km/h}$
(C) $52\text{ km/h}$
(D) $45\text{ km/h}$
Problem 5 (JEE Main – Single Correct):
A particle moves along a straight line such that $x(t) = 8t – 2t^2\text{ m}$. The ratio of average speed to the magnitude of average velocity in the time interval $t = 0$ to $t = 3\text{ s}$ is:
(A) $1.0$
(B) $\frac{5}{3}$
(C) $\frac{4}{3}$
(D) $2.0$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct regarding speed and velocity?
(A) A particle can have constant speed and varying velocity.
(B) A particle can have constant velocity and varying speed.
(C) The magnitude of average velocity is always less than or equal to average speed.
(D) The instantaneous speed is always equal to the magnitude of instantaneous velocity.
Problem 7 (JEE Advanced – One or More Correct):
The displacement of a particle is given by $x(t) = t^3 – 3t\text{ m}$. Which of the following statements is/are correct?
(A) The particle reverses its direction of motion at $t = 1\text{ s}$ (for $t \ge 0$).
(B) The net displacement of the particle from $t = 0$ to $t = 2\text{ s}$ is $2\text{ m}$.
(C) The total distance traveled by the particle from $t = 0$ to $t = 2\text{ s}$ is $6\text{ m}$.
(D) The instantaneous velocity at $t = 2\text{ s}$ is $9\text{ m/s}$.
Problem 8 (JEE Advanced – One or More Correct):
A particle moves along the $x$-axis with velocity $v(t) = 6 – 3t\text{ m/s}$. For the interval $t = 0$ to $t = 4\text{ s}$:
(A) The displacement of the particle is $0\text{ m}$.
(B) The average velocity of the particle is $0\text{ m/s}$.
(C) The total distance covered is $12\text{ m}$.
(D) The average speed is $3\text{ m/s}$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A train completes a journey in three equal time intervals: the first at $30\text{ km/h}$, the second at $45\text{ km/h}$, and the third at $75\text{ km/h}$. The average speed of the train in $\text{km/h}$ is $K$. Find $K$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle moves along the $x$-axis according to $x(t) = 12t – 3t^2\text{ m}$. The average speed in the interval $t = 0$ to $t = 3\text{ s}$ is $V\text{ m/s}$. Find the numerical value of $10 \times V$.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (B) | 3. (B) | 4. (B) | 5. (B) | 6. (A, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 50 | 10. 50
Solution 1:
Total distance $= 2d$. Total time $= \frac{d}{v_1} + \frac{d}{v_2} = \frac{d(v_1 + v_2)}{v_1 v_2}$.
$v_{\text{avg}} = \frac{2d}{\frac{d(v_1 + v_2)}{v_1 v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}$.
Correct Answer: (B)
Solution 2:
$v = \frac{dx}{dt} = 2at – 3bt^2 = 0 \implies t(2a – 3bt) = 0 \implies t = \frac{2a}{3b}$.
Correct Answer: (B)
Solution 3:
In half a revolution, displacement $= 2R$.
Time taken $= \frac{\pi R}{v}$.
Average velocity $= \frac{2R}{\frac{\pi R}{v}} = \frac{2v}{\pi}$.
Correct Answer: (B)
Solution 4:
Equal times $\implies$ arithmetic mean:
$v_{\text{avg}} = \frac{v_1 + v_2}{2} = \frac{40 + 60}{2} = 50\text{ km/h}$.
Correct Answer: (B)
Solution 5:
$x(t) = 8t – 2t^2 \implies v(t) = 8 – 4t = 0$ at $t = 2\text{ s}$.
$x(0) = 0$, $x(2) = 8\text{ m}$, $x(3) = 6\text{ m}$.
Displacement $= x(3) – x(0) = 6\text{ m}$. $|\vec{v}_{\text{avg}}| = \frac{6}{3} = 2\text{ m/s}$.
Distance $= |8 – 0| + |6 – 8| = 8 + 2 = 10\text{ m}$. Average speed $= \frac{10}{3}\text{ m/s}$.
Ratio $= \frac{10/3}{2} = \frac{5}{3}$.
Correct Answer: (B)
Solution 6:
– (A) True (e.g., uniform circular motion has constant speed but continuously changing velocity direction).
– (B) False (constant velocity implies constant speed).
– (C) True ($\text{Distance} \ge |\text{Displacement}| \implies v_{\text{avg}} \ge |\vec{v}_{\text{avg}}|$).
– (D) True ($v_{\text{inst}} = |\vec{v}| = |\frac{d\vec{r}}{dt}|$).
Correct Answer: (A, C, D)
Solution 7:
$v(t) = 3t^2 – 3 = 0 \implies t = 1\text{ s}$ (turning point).
$x(0) = 0$, $x(1) = -2\text{ m}$, $x(2) = 2^3 – 3(2) = 2\text{ m}$.
– Net displacement: $x(2) – x(0) = 2 – 0 = 2\text{ m}$.
– Total distance: $|-2 – 0| + |2 – (-2)| = 2 + 4 = 6\text{ m}$.
– Instantaneous velocity at $t = 2\text{ s}$: $v(2) = 3(2^2) – 3 = 9\text{ m/s}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
$v(t) = 6 – 3t \implies t_{\text{stop}} = 2\text{ s}$.
$x(t) = 6t – 1.5t^2$.
$x(0) = 0$, $x(2) = 6(2) – 1.5(4) = 6\text{ m}$, $x(4) = 6(4) – 1.5(16) = 0\text{ m}$.
– Displacement $= x(4) – x(0) = 0\text{ m}$ (A, B are True).
– Distance $= |6 – 0| + |0 – 6| = 12\text{ m}$ (C is True).
– Average speed $= \frac{12}{4} = 3\text{ m/s}$ (D is True).
All options are correct.
Correct Answer: (A, B, C, D)
Solution 9:
Equal time intervals $\implies$ arithmetic mean:
$v_{\text{avg}} = \frac{30 + 45 + 75}{3} = \frac{150}{3} = 50\text{ km/h}$.
Correct Answer: 50
Solution 10:
$x(t) = 12t – 3t^2 \implies v(t) = 12 – 6t = 0 \implies t = 2\text{ s}$ (turning point).
$x(0) = 0$, $x(2) = 12(2) – 3(4) = 12\text{ m}$, $x(3) = 12(3) – 3(9) = 9\text{ m}$.
Total distance $= |12 – 0| + |9 – 12| = 12 + 3 = 15\text{ m}$.
Average speed $V = \frac{15\text{ m}}{3\text{ s}} = 5\text{ m/s}$.
$10 \times V = 10 \times 5 = 50$.
Correct Answer: 50