Concept Card: Velocity-Time (v-t) Graphs & Graphical Kinematics
1. Physical Significance of Slope in a $v-t$ Graph:
The slope of a velocity-time graph represents the time rate of change of velocity, which is acceleration.
- Instantaneous Acceleration ($a$): Given by the slope of the tangent line drawn to the $v-t$ curve at time $t$:
$a = \frac{dv}{dt} = \tan\theta$
where $\theta$ is the angle made by the tangent line with the positive time axis.- $\theta < 90^\circ \implies \tan\theta > 0 \implies a > 0$ (positive acceleration).
- $\theta = 0^\circ \implies \tan\theta = 0 \implies a = 0$ (uniform velocity / zero acceleration).
- $90^\circ < \theta < 180^\circ \implies \tan\theta < 0 \implies a < 0$ (negative acceleration / retardation when $v > 0$).
- Average Acceleration ($a_{\text{avg}}$): Given by the slope of the secant chord joining two points $(t_1, v_1)$ and $(t_2, v_2)$:
$a_{\text{avg}} = \frac{v_2 – v_1}{t_2 – t_1} = \frac{\Delta v}{\Delta t}$
2. Physical Significance of Area under a $v-t$ Graph:
The area enclosed between the $v-t$ curve, the time axis, and the bounding ordinates $t_1$ and $t_2$ represents the integral of velocity with respect to time.
- Net Displacement ($\Delta x$): The signed area under the $v-t$ graph:
$\Delta x = \int_{t_1}^{t_2} v(t)\,dt = \text{Area above time axis} – \text{Area below time axis}$ - Total Distance Traveled ($s$): The unsigned total area (magnitude sum):
$s = \int_{t_1}^{t_2} |v(t)|\,dt = |\text{Area above time axis}| + |\text{Area below time axis}|$
3. Graphical Derivation of Equations of Motion (Constant Acceleration):
Consider a particle moving with initial velocity $u$ at $t = 0$, accelerating at a constant rate $a$ to acquire velocity $v$ at time $t$. The $v-t$ graph is a straight line sloping upwards from $(0, u)$ to $(t, v)$.
- First Equation ($v = u + at$):
$\text{Slope of line } = \frac{\text{Vertical rise}}{\text{Horizontal run}} = \frac{v – u}{t – 0} = a \implies v – u = at \implies v = u + at$. - Second Equation ($s = ut + \frac{1}{2}at^2$):
Displacement $s$ is the area of the trapezoid under the $v-t$ line from $0$ to $t$:
$s = \text{Area of rectangle } (u \times t) + \text{Area of upper triangle } \left(\frac{1}{2} \times t \times (v – u)\right)$
Substitute $(v – u) = at$:
$s = ut + \frac{1}{2}t(at) = ut + \frac{1}{2}at^2$. - Third Equation ($v^2 = u^2 + 2as$):
Express the area of the trapezoid as $\frac{1}{2}(\text{sum of parallel sides}) \times \text{height}$:
$s = \frac{u + v}{2} \cdot t$
From the first equation, substitute $t = \frac{v – u}{a}$:
$s = \left(\frac{v + u}{2}\right)\left(\frac{v – u}{a}\right) = \frac{v^2 – u^2}{2a} \implies v^2 = u^2 + 2as$.
Solved Examples
Example 1 (Distance and Displacement from Signed Area):
The velocity-time graph of a particle consists of three segments:
– From $t = 0$ to $t = 4\text{ s}$, velocity increases linearly from $0$ to $+10\text{ m/s}$.
– From $t = 4\text{ s}$ to $t = 8\text{ s}$, velocity is constant at $+10\text{ m/s}$.
– From $t = 8\text{ s}$ to $t = 12\text{ s}$, velocity decreases linearly from $+10\text{ m/s}$ to $-10\text{ m/s}$ (crossing $v = 0$ at $t = 10\text{ s}$).
Calculate:
(a) The total displacement of the particle over $12\text{ s}$.
(b) The total distance covered by the particle over $12\text{ s}$.
(c) The average velocity and average speed.
Solution:
1. Segment areas above the time axis ($t \in [0, 10\text{ s}]$):
– Triangle 1 ($0 \to 4\text{ s}$): $A_1 = \frac{1}{2} \times 4 \times 10 = 20\text{ m}$.
– Rectangle 2 ($4 \to 8\text{ s}$): $A_2 = (8 – 4) \times 10 = 40\text{ m}$.
– Triangle 3 ($8 \to 10\text{ s}$): $A_3 = \frac{1}{2} \times (10 – 8) \times 10 = 10\text{ m}$.
Total positive area: $A_+ = 20 + 40 + 10 = 70\text{ m}$.
2. Segment area below the time axis ($t \in [10, 12\text{ s}]$):
– Triangle 4 ($10 \to 12\text{ s}$): $A_- = \frac{1}{2} \times (12 – 10) \times (-10) = -10\text{ m}$.
(a) Net displacement: $\Delta x = A_+ + A_- = 70 – 10 = 60\text{ m}$.
(b) Total distance: $s = |A_+| + |A_-| = 70 + 10 = 80\text{ m}$.
(c) Average velocity: $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{60\text{ m}}{12\text{ s}} = 5.0\text{ m/s}$.
Average speed: $\text{Speed}_{\text{avg}} = \frac{s}{\Delta t} = \frac{80\text{ m}}{12\text{ s}} \approx 6.67\text{ m/s}$.
Example 2 (Trapezoidal Motion Profile – Metro Rail Transit):
An electric train accelerates from rest at rate $\alpha = 1.5\text{ m/s}^2$ until it reaches cruising speed $v_{\text{cruise}}$, travels at constant speed for time $t_2$, and then decelerates at rate $\beta = 3.0\text{ m/s}^2$ to come to a stop at the next station. The total distance is $S = 1800\text{ m}$ and total transit time is $T = 75\text{ s}$. Find the cruising speed $v_{\text{cruise}}$.
Solution:
1. Time taken during acceleration: $t_1 = \frac{v_{\text{cruise}}}{\alpha} = \frac{v_{\text{cruise}}}{1.5} = \frac{2}{3}v_{\text{cruise}}$.
2. Time taken during deceleration: $t_3 = \frac{v_{\text{cruise}}}{\beta} = \frac{v_{\text{cruise}}}{3.0} = \frac{1}{3}v_{\text{cruise}}$.
3. Total acceleration + deceleration time: $t_1 + t_3 = \left(\frac{2}{3} + \frac{1}{3}\right)v_{\text{cruise}} = v_{\text{cruise}}$.
4. Constant velocity duration: $t_2 = T – (t_1 + t_3) = 75 – v_{\text{cruise}}$.
5. Total distance is the area of the trapezoid:
$S = \frac{1}{2}(\text{Base}_1 + \text{Base}_2) \times \text{Height} = \frac{1}{2}(T + t_2) \cdot v_{\text{cruise}}$.
Substitute $S = 1800$, $T = 75$, and $t_2 = 75 – v_{\text{cruise}}$:
$1800 = \frac{1}{2} [75 + (75 – v_{\text{cruise}})] \cdot v_{\text{cruise}} = \frac{1}{2} (150 – v_{\text{cruise}}) \cdot v_{\text{cruise}}$
$3600 = 150 v_{\text{cruise}} – v_{\text{cruise}}^2 \implies v_{\text{cruise}}^2 – 150 v_{\text{cruise}} + 3600 = 0$.
Factoring:
$(v_{\text{cruise}} – 30)(v_{\text{cruise}} – 120) = 0$.
Since $v_{\text{cruise}} = 120\text{ m/s}$ would make $t_1 + t_3 = 120\text{ s} > 75\text{ s}$, that root is physically unfeasible.
Therefore: $v_{\text{cruise}} = 30\text{ m/s}$ ($108\text{ km/h}$).
Example 3 (Two Vehicles Overtaking Analyzed via $v-t$ Graph):
At $t = 0$, Car $A$ is moving with constant velocity $v_A = 20\text{ m/s}$. At the same instant and position, Car $B$ starts from rest and accelerates with constant acceleration $a_B = 2.5\text{ m/s}^2$.
(a) Draw the qualitative $v-t$ graph and determine the time when Car $B$ matches Car $A$’s speed.
(b) Find the time when Car $B$ overtakes Car $A$.
(c) Find the maximum lead Car $A$ maintains over Car $B$ before being overtaken.
Solution:
(a) Velocity of $B$: $v_B(t) = a_B t = 2.5 t$.
Velocities match when $v_B = v_A \implies 2.5 t = 20 \implies t = 8\text{ seconds}$.
(b) Overtaking occurs when the area under both $v-t$ curves from $t = 0$ to $t$ are equal (equal displacement):
$s_A = v_A \cdot t = 20 t$.
$s_B = \frac{1}{2} a_B t^2 = \frac{1}{2}(2.5) t^2 = 1.25 t^2$.
Equating areas: $1.25 t^2 = 20 t \implies t = \frac{20}{1.25} = 16\text{ seconds}$.
(c) The separation $\Delta x = s_A – s_B$ is maximum when the relative velocity is zero ($v_A = v_B$), which occurs at $t = 8\text{ s}$:
$\Delta x_{\text{max}} = s_A(8) – s_B(8) = 20(8) – \frac{1}{2}(2.5)(8^2) = 160 – 80 = 80\text{ m}$.
Geometrical interpretation: The maximum separation is the area of the triangle enclosed between the horizontal line of Car $A$ and the ramp of Car $B$ from $t = 0$ to $t = 8\text{ s}$: $\frac{1}{2} \times 8 \times 20 = 80\text{ m}$.
Example 4 (Curved $v-t$ Graph with Calculus Integration):
The velocity of a particle moving along the $x$-axis varies with time as $v(t) = 3t^2 – 12t\text{ m/s}$ for $t \ge 0$.
(a) Find the instantaneous acceleration at $t = 3\text{ s}$.
(b) Find the time when the particle is momentarily at rest (other than $t = 0$).
(c) Find the total distance and net displacement from $t = 0$ to $t = 6\text{ s}$.
Solution:
(a) Slope of $v-t$ curve gives acceleration:
$a(t) = \frac{dv}{dt} = 6t – 12\text{ m/s}^2$.
At $t = 3\text{ s}$: $a(3) = 6(3) – 12 = +6\text{ m/s}^2$.
(b) Momentary rest: $v(t) = 0 \implies 3t(t – 4) = 0 \implies t = 4\text{ seconds}$.
(c) For $t \in [0, 4\text{ s}]$, $v(t) \le 0$ (area is below time axis).
$A_1 = \int_{0}^{4} (3t^2 – 12t)\,dt = [t^3 – 6t^2]_{0}^{4} = 64 – 96 = -32\text{ m}$.
For $t \in [4, 6\text{ s}]$, $v(t) \ge 0$ (area is above time axis).
$A_2 = \int_{4}^{6} (3t^2 – 12t)\,dt = [t^3 – 6t^2]_{4}^{6} = (216 – 216) – (64 – 96) = 0 – (-32) = +32\text{ m}$.
– Net displacement: $\Delta x = A_1 + A_2 = -32 + 32 = 0\text{ m}$.
– Total distance: $s = |A_1| + |A_2| = 32 + 32 = 64\text{ m}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The velocity-time graph of an object is an equilateral triangle in the upper half-plane with a base of $10\text{ s}$ on the time axis and a peak height of $20\text{ m/s}$. The total displacement of the object is:
(A) $100\text{ m}$
(B) $200\text{ m}$
(C) $50\text{ m}$
(D) $150\text{ m}$
Problem 2 (JEE Main – Single Correct):
The slope of a velocity-time graph at any given instant yields:
(A) Displacement
(B) Distance
(C) Instantaneous acceleration
(D) Average speed
Problem 3 (JEE Main – Single Correct):
A ball is thrown vertically upwards with velocity $u$, reaches maximum height, and falls back to the thrower’s hand. Which of the following best describes its $v-t$ graph (taking upward direction as positive)?
(A) A horizontal straight line above the time axis.
(B) A straight line with a constant negative slope passing continuously through the time axis.
(C) A parabola opening upwards.
(D) A straight line with a positive slope.
Problem 4 (JEE Main – Single Correct):
The area bounded by the velocity-time graph and the time axis between two time instants represents:
(A) Acceleration
(B) Change in position (Displacement)
(C) Change in acceleration
(D) Force
Problem 5 (JEE Main – Single Correct):
The $v-t$ graph of an elevator ascending from the ground floor to the top floor of a skyscraper is a trapezoid: it accelerates at $2\text{ m/s}^2$ for $4\text{ s}$, moves at constant speed for $10\text{ s}$, and decelerates to rest at $2\text{ m/s}^2$ for $4\text{ s}$. The height of the skyscraper floor reached is:
(A) $96\text{ m}$
(B) $112\text{ m}$
(C) $128\text{ m}$
(D) $144\text{ m}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding velocity-time ($v-t$) graphs?
(A) The area under the $v-t$ curve can be negative, but the area under the $|v|-t$ curve is always non-negative.
(B) If the $v-t$ graph is symmetric about the time axis, the net displacement over that interval is zero.
(C) A point where the $v-t$ curve crosses the time axis represents a reversal of the direction of motion.
(D) The average acceleration over an interval equals the slope of the secant connecting the interval’s endpoints.
Problem 7 (JEE Advanced – One or More Correct):
A particle moves along a straight line. Its $v-t$ curve is given by $v(t) = 4t – t^2\text{ m/s}$ for $t \in [0, 6\text{ s}]$. Which of the following statements is/are correct?
(A) The particle attains maximum velocity at $t = 2\text{ s}$.
(B) The particle reverses its direction of motion at $t = 4\text{ s}$.
(C) The net displacement between $t = 0$ and $t = 6\text{ s}$ is $0\text{ m}$.
(D) The total distance traveled between $t = 0$ and $t = 6\text{ s}$ is $\frac{64}{3}\text{ m}$.
Problem 8 (JEE Advanced – One or More Correct):
Two particles $A$ and $B$ start simultaneously from the same point along the same straight line. Their $v-t$ graphs are shown such that $v_A(t) = 10\text{ m/s}$ (constant) and $v_B(t) = 2t\text{ m/s}$. Which statements are TRUE?
(A) At $t = 5\text{ s}$, both particles have equal velocity.
(B) At $t = 5\text{ s}$, the separation between the particles is maximum.
(C) At $t = 10\text{ s}$, particle $B$ overtakes particle $A$.
(D) The maximum lead of particle $A$ over particle $B$ is $25\text{ m}$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A car starts from rest and accelerates uniformly at rate $\alpha = 4\text{ m/s}^2$ for time $t_1$, then immediately decelerates uniformly at rate $\beta = 2\text{ m/s}^2$ to come to rest in total time $T = 15\text{ s}$. The total distance covered by the car is $S\text{ meters}$. Find $S$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
The $v-t$ graph of a particle is a triangle of base $6\text{ s}$ and height $12\text{ m/s}$ above the time axis, followed by a triangle of base $4\text{ s}$ and depth $-6\text{ m/s}$ below the time axis. The ratio of total distance to magnitude of net displacement is $R$. Find the numerical value of $10 \times R$.
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, C, D) | 9. 150 | 10. 20
Solution 1:
Displacement is the area of the triangle:
$\text{Displacement} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10\text{ s} \times 20\text{ m/s} = 100\text{ m}$.
Correct Answer: (A)
Solution 2:
By definition: $\text{Slope} = \frac{dv}{dt} = \text{instantaneous acceleration}$.
Correct Answer: (C)
Solution 3:
For vertical projectile motion, acceleration is strictly constant ($a = -g$).
$v(t) = u – gt$. This represents a straight line with initial velocity $+u$ and constant negative slope $-g$ that crosses zero at peak height and becomes negative.
Correct Answer: (B)
Solution 4:
$\int v\,dt = \Delta x = \text{Displacement}$.
Correct Answer: (B)
Solution 5:
Cruising speed: $v_{\text{max}} = a \cdot t_1 = 2 \times 4 = 8\text{ m/s}$.
Total time: $T = 4 + 10 + 4 = 18\text{ s}$.
Flat top duration: $t_{\text{flat}} = 10\text{ s}$.
Height is the area of the trapezoid:
$h = \frac{1}{2}(T + t_{\text{flat}}) \cdot v_{\text{max}} = \frac{1}{2}(18 + 10) \times 8 = 14 \times 8 = 112\text{ m}$.
Correct Answer: (B)
Solution 6:
All four statements (A, B, C, D) are fundamental and mathematically rigorous properties of velocity-time graphs in kinematics.
Correct Answer: (A, B, C, D)
Solution 7:
$v(t) = 4t – t^2$.
– Maximum velocity when $\frac{dv}{dt} = 4 – 2t = 0 \implies t = 2\text{ s}$ (A is True).
– Direction reverses when $v(t) = 0 \implies t(4 – t) = 0 \implies t = 4\text{ s}$ (B is True).
– Net displacement: $\int_{0}^{6} (4t – t^2)\,dt = [2t^2 – \frac{t^3}{3}]_{0}^{6} = 2(36) – \frac{216}{3} = 72 – 72 = 0\text{ m}$ (C is True).
– Positive area ($0 \to 4\text{ s}$): $[2t^2 – \frac{t^3}{3}]_{0}^{4} = 32 – \frac{64}{3} = \frac{32}{3}\text{ m}$.
– Negative area ($4 \to 6\text{ s}$): $|[2t^2 – \frac{t^3}{3}]_{4}^{6}| = |0 – \frac{32}{3}| = \frac{32}{3}\text{ m}$.
– Total distance: $\frac{32}{3} + \frac{32}{3} = \frac{64}{3}\text{ m}$ (D is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– Velocities match when $2t = 10 \implies t = 5\text{ s}$ (A is True).
– Separation $\Delta x = 10t – t^2$ is maximum when relative velocity is zero, which is $t = 5\text{ s}$ (B is True).
– Displacements match when $10t = t^2 \implies t = 10\text{ s}$ (C is True).
– Maximum separation: $\Delta x_{\text{max}} = 10(5) – 5^2 = 50 – 25 = 25\text{ m}$ (D is True).
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 9:
For acceleration $\alpha$ and deceleration $\beta$ with total time $T$:
$S = \frac{1}{2}\left(\frac{\alpha\beta}{\alpha + \beta}\right) T^2$.
$S = \frac{1}{2}\left(\frac{4 \times 2}{4 + 2}\right) \times 15^2 = \frac{1}{2}\left(\frac{8}{6}\right) \times 225 = \frac{2}{3} \times 225 = 150\text{ meters}$.
Correct Answer: 150
Solution 10:
– Area above: $A_1 = \frac{1}{2} \times 6 \times 12 = 36\text{ m}$.
– Area below: $A_2 = \frac{1}{2} \times 4 \times (-6) = -12\text{ m}$.
– Displacement: $\Delta x = 36 – 12 = 24\text{ m}$.
– Distance: $s = 36 + 12 = 48\text{ m}$.
– Ratio: $R = \frac{s}{|\Delta x|} = \frac{48}{24} = 2.0$.
$10 \times R = 10 \times 2.0 = 20$.
Correct Answer: 20