Concept Card: Position-Time (x-t) Graphs & Motion Interpretation
1. Physical Meaning of the Slope of an $x-t$ Graph:
The slope of a position-time graph represents the rate of change of position with respect to time, which is velocity.
- Instantaneous Velocity ($v$): The slope of the tangent line drawn to the $x-t$ curve at any instant $t$:
$v = \frac{dx}{dt} = \tan\theta$
where $\theta$ is the angle made by the tangent line with the positive time ($t$) axis.- $\theta < 90^\circ \implies \tan\theta > 0 \implies v > 0$ (particle moving in the positive $x$-direction).
- $\theta = 0^\circ \implies \tan\theta = 0 \implies v = 0$ (particle is instantaneously at rest; local maximum, local minimum, or horizontal line).
- $90^\circ < \theta < 180^\circ \implies \tan\theta < 0 \implies v < 0$ (particle moving in the negative $x$-direction).
- $\theta = 90^\circ \implies \tan\theta \to \infty$ (infinite speed – physically impossible for any mass).
- Average Velocity ($v_{\text{avg}}$): The slope of the secant (chord) line connecting two points $(t_1, x_1)$ and $(t_2, x_2)$ on the curve:
$v_{\text{avg}} = \frac{x_2 – x_1}{t_2 – t_1} = \frac{\Delta x}{\Delta t}$
2. Physical Meaning of Curvature (Concavity) and Acceleration:
The second derivative of position with respect to time yields acceleration: $a = \frac{d^2 x}{dt^2} = \frac{d}{dt}\left(\frac{dx}{dt}\right)$.
- Concave Upwards ($\cup$ shape, tangent rotates counterclockwise):
$\frac{d^2 x}{dt^2} > 0 \implies a > 0$ (positive acceleration).
Speed increases if $v > 0$; speed decreases if $v < 0$. - Concave Downwards ($\cap$ shape, tangent rotates clockwise):
$\frac{d^2 x}{dt^2} < 0 \implies a < 0$ (negative acceleration / retardation when $v > 0$).
Speed decreases if $v > 0$; speed increases in negative direction if $v < 0$. - Zero Curvature (Straight Line):
$\frac{d^2 x}{dt^2} = 0 \implies a = 0$ (uniform motion with constant velocity). - Point of Inflection: A point where the curve changes concavity ($\frac{d^2 x}{dt^2} = 0$ while crossing zero). At this instant, acceleration is momentarily zero ($a = 0$).
3. Physical Criteria & Impossible $x-t$ Graphs:
- Multi-Valued Position: A single time $t$ can never have more than one corresponding position $x$ (e.g., closed circular or vertical loops are physically impossible because a particle cannot exist at two different places at the same instant).
- Time Invariance: Time can never reverse or flow backward ($\frac{dt}{dx}$ cannot be negative).
- Cusps / Sharp Corners: A sharp discontinuity in slope implies an instantaneous jump in velocity ($\Delta v$ in $\Delta t = 0$), requiring an infinite acceleration ($a \to \infty$) and infinite force, which is physically unrealizable in classical mechanics.
Solved Examples
Example 1 (Segmented Piecewise Linear $x-t$ Graph):
An object moves along the $x$-axis such that its $x-t$ graph consists of three continuous linear segments:
– Segment 1: From $t = 0$ to $t = 4\text{ s}$, $x$ increases linearly from $0$ to $20\text{ m}$.
– Segment 2: From $t = 4\text{ s}$ to $t = 8\text{ s}$, $x$ remains constant at $20\text{ m}$.
– Segment 3: From $t = 8\text{ s}$ to $t = 12\text{ s}$, $x$ decreases linearly from $20\text{ m}$ to $-10\text{ m}$.
Calculate:
(a) The velocity in each segment.
(b) The net displacement over the entire $12\text{ s}$.
(c) The total distance traveled.
(d) The average velocity and average speed from $t = 0$ to $t = 12\text{ s}$.
Solution:
(a) Slope gives velocity in each linear section:
– $v_1 = \frac{20 – 0}{4 – 0} = +5.0\text{ m/s}$.
– $v_2 = \frac{20 – 20}{8 – 4} = 0\text{ m/s}$ (object is at rest).
– $v_3 = \frac{-10 – 20}{12 – 8} = \frac{-30}{4} = -7.5\text{ m/s}$ (moving in negative $x$-direction).
(b) Net displacement: $\Delta x = x(12) – x(0) = -10 – 0 = -10\text{ m}$.
(c) Distance traveled is the sum of absolute displacements of segments:
$s = |20 – 0| + |20 – 20| + |-10 – 20| = 20 + 0 + 30 = 50\text{ m}$.
(d) Average velocity: $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{-10\text{ m}}{12\text{ s}} \approx -0.833\text{ m/s}$.
Average speed: $\text{Speed}_{\text{avg}} = \frac{s}{\Delta t} = \frac{50\text{ m}}{12\text{ s}} \approx 4.17\text{ m/s}$.
Example 2 (Parabolic Position-Time Curve & Extrema):
The position of a particle is described by the parabolic equation $x(t) = -t^2 + 6t + 7\text{ m}$ for $t \ge 0$.
(a) Determine the initial position and initial velocity.
(b) Find the time and position where the slope of the $x-t$ curve is zero.
(c) Interpret the curvature of the graph and find acceleration.
(d) Compute the total distance covered from $t = 0$ to $t = 5\text{ s}$.
Solution:
(a) At $t = 0$: $x(0) = 0 + 0 + 7 = +7\text{ m}$.
Velocity: $v(t) = \frac{dx}{dt} = -2t + 6\text{ m/s}$.
Initial velocity: $v(0) = +6\text{ m/s}$.
(b) Slope is zero when $v(t) = 0 \implies -2t + 6 = 0 \implies t = 3\text{ s}$.
Position at turning point: $x(3) = -(3)^2 + 6(3) + 7 = -9 + 18 + 7 = +16\text{ m}$ (crest of the parabola).
(c) Acceleration: $a = \frac{d^2 x}{dt^2} = -2\text{ m/s}^2$.
Since $a = -2 < 0$, the parabola is concave downwards ($\cap$) everywhere, indicating constant negative acceleration.
(d) Position at $t = 5\text{ s}$: $x(5) = -(5)^2 + 6(5) + 7 = -25 + 30 + 7 = +12\text{ m}$.
Distance from $t = 0$ to $t = 3\text{ s}$: $|x(3) – x(0)| = |16 – 7| = 9\text{ m}$.
Distance from $t = 3$ to $t = 5\text{ s}$: $|x(5) – x(3)| = |12 – 16| = 4\text{ m}$.
Total distance: $s = 9 + 4 = 13\text{ m}$.
Example 3 (Intersection of Two Motion Graphs & Collision):
Car $A$ starts from the origin ($x_A = 0$) at $t = 0$ and moves with constant velocity $v_A = 4\text{ m/s}$. Car $B$ starts from $x_B = 16\text{ m}$ at rest ($u_B = 0$) at $t = 0$ and moves towards the origin with constant acceleration $a_B = 2\text{ m/s}^2$ ($a = -2\text{ m/s}^2$).
(a) Write the equations of their position-time graphs.
(b) Find the time and position where their graphs intersect (meeting point).
(c) Find the relative velocity between Car $A$ and Car $B$ at the intersection.
Solution:
(a) For Car $A$ (straight line): $x_A(t) = 0 + 4t = 4t$.
For Car $B$ (parabola opening downwards): $x_B(t) = 16 + 0\cdot t – \frac{1}{2}(2)t^2 = 16 – t^2$.
(b) Intersection occurs when $x_A(t) = x_B(t)$:
$4t = 16 – t^2 \implies t^2 + 4t – 16 = 0$.
Solving using quadratic formula for $t > 0$:
$t = \frac{-4 \pm \sqrt{16 – 4(1)(-16)}}{2} = \frac{-4 \pm \sqrt{80}}{2} = -2 + 2\sqrt{5}\text{ s} \approx -2 + 4.472 = 2.472\text{ s}$.
Intersection position: $x = 4(2.472) \approx 9.89\text{ m}$.
(c) Velocities at intersection ($t \approx 2.472\text{ s}$):
$v_A = +4\text{ m/s}$.
$v_B = \frac{dx_B}{dt} = -2t = -2(2.472) = -4.944\text{ m/s}$.
Relative velocity of $A$ with respect to $B$:
$v_{AB} = v_A – v_B = 4 – (-4.944) = +8.944\text{ m/s}$.
Example 4 (Ratio of Velocities from Inclination Angles):
The position-time graphs of two particles $P$ and $Q$ moving along parallel lines are straight lines inclined at angles $\theta_P = 30^\circ$ and $\theta_Q = 60^\circ$ with the time axis.
(a) Calculate the ratio of their velocities $\frac{v_P}{v_Q}$.
(b) If both particles started from the origin, but particle $Q$ started $2\text{ seconds}$ after particle $P$, find the time $t$ elapsed from the start of $P$ when $Q$ overtakes $P$.
Solution:
(a) Velocities are given by slopes:
$v_P = \tan(30^\circ) = \frac{1}{\sqrt{3}}\text{ m/s}$.
$v_Q = \tan(60^\circ) = \sqrt{3}\text{ m/s}$.
Ratio: $\frac{v_P}{v_Q} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.
(b) Equations of motion:
$x_P(t) = v_P \cdot t = \frac{1}{\sqrt{3}} t$ (for $t \ge 0$).
$x_Q(t) = v_Q \cdot (t – 2) = \sqrt{3}(t – 2)$ (for $t \ge 2\text{ s}$).
Overtaking occurs when $x_Q = x_P$:
$\sqrt{3}(t – 2) = \frac{1}{\sqrt{3}} t \implies 3(t – 2) = t \implies 3t – 6 = t \implies 2t = 6 \implies t = 3\text{ seconds}$.
Particle $Q$ overtakes particle $P$ at $t = 3\text{ s}$ from the start of $P$ (which is $1\text{ s}$ after $Q$ starts).
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The position-time graphs of two bodies $A$ and $B$ are straight lines making angles of $30^\circ$ and $45^\circ$ with the time axis respectively. The ratio of their velocities $\frac{v_A}{v_B}$ is:
(A) $\sqrt{3} : 1$
(B) $1 : \sqrt{3}$
(C) $1 : 3$
(D) $3 : 1$
Problem 2 (JEE Main – Single Correct):
A body moves along a straight line. If its position-time graph is a parabola opening downwards ($\cap$), the acceleration of the body is:
(A) Positive and increasing
(B) Positive and constant
(C) Negative and constant
(D) Zero
Problem 3 (JEE Main – Single Correct):
Which of the following $x-t$ graph features represents a physically IMPOSSIBLE situation in classical mechanics?
(A) An $x-t$ curve having a horizontal tangent at some point.
(B) An $x-t$ curve having a vertical tangent at some point.
(C) An $x-t$ curve passing through the origin.
(D) An $x-t$ curve entering the negative $x$ region.
Problem 4 (JEE Main – Single Correct):
The displacement-time equation of a particle is given by $x(t) = 4t – t^2\text{ m}$. The particle reverses its direction of motion at time $t$ equal to:
(A) $1\text{ s}$
(B) $2\text{ s}$
(C) $4\text{ s}$
(D) $0.5\text{ s}$
Problem 5 (JEE Main – Single Correct):
A particle moves along the $x$-axis. Its $x-t$ graph is a semicircle of radius $R = 5\text{ m}$ centered at $(t = 5\text{ s}, x = 0)$ in the upper half-plane. The velocity of the particle at $t = 5\text{ s}$ is:
(A) $0\text{ m/s}$
(B) $1\text{ m/s}$
(C) $5\text{ m/s}$
(D) Undefined / Infinite
Problem 6 (JEE Advanced – One or More Correct):
The position-time graph of a particle is given by $x(t) = t^3 – 6t^2 + 9t\text{ m}$ for $t \ge 0$. Which of the following statements is/are correct?
(A) The particle comes to instantaneous rest at $t = 1\text{ s}$ and $t = 3\text{ s}$.
(B) The acceleration is zero at $t = 2\text{ s}$ (point of inflection).
(C) The particle speeds up in the interval $t = 1\text{ s}$ to $t = 2\text{ s}$.
(D) The net displacement between $t = 0$ and $t = 3\text{ s}$ is $0\text{ m}$.
Problem 7 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding position-time ($x-t$) graphs?
(A) The area under an $x-t$ graph has no direct physical meaning in kinematics.
(B) A straight line parallel to the position axis represents instantaneous teleportation / infinite speed.
(C) If the slope of an $x-t$ curve is non-zero and constant, the particle has zero acceleration.
(D) The average velocity over an interval $[t_1, t_2]$ equals the slope of the chord joining $(t_1, x_1)$ and $(t_2, x_2)$.
Problem 8 (JEE Advanced – One or More Correct):
Two cars $A$ and $B$ are traveling along a straight road. Their position-time curves intersect at point $(t_0, x_0)$. At this point, the tangent to curve $A$ has a greater slope than the tangent to curve $B$. Which of the following statements is/are correct?
(A) Car $A$ and Car $B$ are at the same location at time $t_0$.
(B) At time $t_0$, Car $A$ is moving faster than Car $B$.
(C) Car $A$ is overtaking Car $B$ at time $t_0$.
(D) Car $A$ and Car $B$ have equal acceleration at time $t_0$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
An object moves along the $x$-axis. From $t = 0$ to $t = 5\text{ s}$, its position increases linearly from $x = 0$ to $x = 30\text{ m}$. From $t = 5\text{ s}$ to $t = 10\text{ s}$, its position decreases linearly from $x = 30\text{ m}$ to $x = 0$. The average speed of the object over the entire $10\text{ seconds}$ is $V\text{ m/s}$. Find the value of $V$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
The position-time graph of a body under constant deceleration is given by $x(t) = 16t – 2t^2\text{ m}$. The maximum coordinate (peak of the curve) reached by the particle before reversing direction is $X_{\text{max}}\text{ meters}$. Determine the integer value of $X_{\text{max}}$.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (C) | 3. (B) | 4. (B) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 6 | 10. 32
Solution 1:
Velocity is the slope of the $x-t$ graph:
$v_A = \tan(30^\circ) = \frac{1}{\sqrt{3}}$
$v_B = \tan(45^\circ) = 1$
Ratio $\frac{v_A}{v_B} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}} = 1 : \sqrt{3}$.
Correct Answer: (B)
Solution 2:
A parabola has the general equation $x(t) = x_0 + ut + \frac{1}{2}at^2$.
For a parabola opening downwards, the coefficient of $t^2$ is negative: $\frac{1}{2}a < 0 \implies a < 0$.
Since it is a parabola, $a = \frac{d^2 x}{dt^2}$ is a constant. Thus acceleration is negative and constant.
Correct Answer: (C)
Solution 3:
A vertical tangent means $\theta = 90^\circ \implies \text{slope} = \frac{dx}{dt} \to \infty$. Infinite velocity violates classical and relativistic physics, and implies zero time for finite displacement. Hence (B) is impossible.
Correct Answer: (B)
Solution 4:
Reversal of direction occurs when velocity is zero: $v = \frac{dx}{dt} = 0$.
$x(t) = 4t – t^2 \implies v(t) = 4 – 2t = 0 \implies t = 2\text{ s}$.
Correct Answer: (B)
Solution 5:
Equation of the semicircle centered at $(5, 0)$ with radius $R = 5$:
$(t – 5)^2 + x^2 = 5^2 = 25 \implies x(t) = \sqrt{25 – (t – 5)^2}$.
At $t = 5\text{ s}$, $x = \sqrt{25 – 0} = 5\text{ m}$ (the highest point of the semicircle).
The tangent at the peak of a circle is perfectly horizontal ($\theta = 0^\circ$).
Therefore, velocity $v = \text{slope} = 0\text{ m/s}$.
Correct Answer: (A)
Solution 6:
$x(t) = t^3 – 6t^2 + 9t$.
$v(t) = \frac{dx}{dt} = 3t^2 – 12t + 9 = 3(t – 1)(t – 3)$.
– $v(t) = 0$ at $t = 1\text{ s}$ and $t = 3\text{ s}$ (A is True).
– $a(t) = \frac{dv}{dt} = 6t – 12 = 0 \implies t = 2\text{ s}$ (B is True).
– For $t \in (1, 2)$: $v < 0$ and $a = 6t - 12 < 0$. Since velocity and acceleration have the same sign (both negative), the particle is speeding up (C is True).
– At $t = 0$: $x(0) = 0$. At $t = 3$: $x(3) = 27 – 54 + 27 = 0$. Net displacement $\Delta x = 0 – 0 = 0\text{ m}$ (D is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: $\int x\,dt$ has units $\text{m}\cdot\text{s}$, which corresponds to absement (no standard kinematic utility).
– (B) True: A vertical line represents infinite velocity.
– (C) True: Constant non-zero slope means constant velocity, so $a = \frac{dv}{dt} = 0$.
– (D) True: Definition of secant slope representing average velocity.
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– Intersection at $(t_0, x_0)$ means both cars share the same position at $t_0$ (A is True).
– Tangent slope gives instantaneous speed. Slope of $A >$ Slope of $B \implies v_A > v_B$ (B is True).
– Since Car $A$ is at the same point with higher speed, it is overtaking Car $B$ (C is True).
– Curvatures (accelerations) are independent of tangent slopes at a single intersection (D is False).
Correct Answer: (A, B, C)
Solution 9:
Total distance $s = |30 – 0| + |0 – 30| = 30 + 30 = 60\text{ m}$.
Total time $\Delta t = 10\text{ s}$.
Average speed $V = \frac{60\text{ m}}{10\text{ s}} = 6\text{ m/s}$.
Correct Answer: 6
Solution 10:
$x(t) = 16t – 2t^2 \implies v(t) = 16 – 4t = 0 \implies t = 4\text{ s}$.
Maximum coordinate: $X_{\text{max}} = x(4) = 16(4) – 2(4^2) = 64 – 32 = 32\text{ m}$.
Correct Answer: 32