Notes: Motion in a Straight Line — Graphs, Concepts, and Problems (Lecture Series)
Kinematics in a Straight Line — Graphs, Concepts, and Problems (Lecture Notes)
Core topics covered- Kinematics in one dimension: the branch of classical mechanics that describes the motion of points, bodies, and systems of bodies without considering the causes of their motion.
Graphs: detailed analysis of displacement–time (x–t), velocity–time (v–t), and acceleration–time (a–t) graphs, including their shapes, slopes, and areas.
Graph-conversion and interpretation: comprehensive methods to read, interpret, and accurately convert between x–t, v–t, and a–t graphs, understanding the underlying mathematical relationships.
Problems on motion under gravity (free fall and variable acceleration): in-depth problem-solving for scenarios involving constant gravitational acceleration and cases where acceleration varies with time, position, or velocity.
Applications to real exam-style questions with solution strategies: practical application of concepts and formulas to solve complex problems, focusing on systematic approaches.
Key concepts and definitions
Displacement vs. Distance
Displacement: change in position of an object; a vector quantity (has both magnitude and direction). It's the shortest distance from the initial to the final position. The displacement can be positive, negative, or zero.
Formula: \text{displacement} = \mathbf{r}f - \mathbf{r}i = \Delta \mathbf{r}
Distance: total length of the path traveled by an object; a scalar quantity (has magnitude only). Distance is always nonnegative.
Speed vs. Velocity
Speed: rate at which an object covers distance; scalar; magnitude only. It indicates how fast an object is moving.
Average Speed: \text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}
Instantaneous Speed: The speed at a specific instant in time, which is the magnitude of the instantaneous velocity.
Velocity: rate at which an object changes its displacement; vector; includes both magnitude (speed) and direction. It indicates both how fast and in what direction an object is moving.
Average Velocity: \text{average velocity} = \frac{\Delta \text{displacement}}{\Delta \text{time}} = \frac{x2-x1}{t2-t1}
Instantaneous Velocity: The limiting value of the average velocity as the time interval approaches zero. Mathematically, v(t) = \frac{dx}{dt}. It's the slope of the x-t graph at a given instant.
Acceleration
Rate of change of velocity with time.
Scalar or vector? Acceleration is a vector quantity; its sign indicates the direction of acceleration. Positive acceleration means velocity is increasing (if moving in the positive direction) or decreasing (if moving in the negative direction, i.e., slowing down). Negative acceleration means velocity is decreasing (if moving in the positive direction, i.e., slowing down) or increasing (if moving in the negative direction, i.e., speeding up in the negative direction).
Average Acceleration: \mathbf{a}_{avg} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{\mathbf{v}-\mathbf{u}}{t} (where u is initial velocity, v is final velocity, t is the time interval).
Instantaneous Acceleration: The limiting value of the average acceleration as the time interval approaches zero. Mathematically, a(t) = \frac{dv}{dt}. It's the slope of the v-t graph at a given instant.
Units (SI)
Distance: meters (m)
Displacement: meters (m)
Speed/Velocity: meters per second (m/s or ms⁻¹)
Acceleration: meters per second squared (m/s² or ms⁻²)
Graph basics
x–t graph (Displacement-Time Graph):
Slope = instantaneous velocity v(t). A positive slope indicates motion in the positive direction; a negative slope indicates motion in the negative direction. A zero slope means the object is at rest.
A straight line indicates constant velocity (zero acceleration).
A curved line indicates changing velocity (non-zero acceleration). A parabolic curve typically indicates constant acceleration.
Concavity: Upward concavity means positive acceleration; downward concavity means negative acceleration.
v–t graph (Velocity-Time Graph):
Slope = instantaneous acceleration a(t). A constant slope implies constant acceleration. A zero slope means constant velocity (zero acceleration).
Area under the curve of v–t (between time t1 and t2) represents the displacement during that interval. The area can be positive or negative depending on the velocity's sign.
a–t graph (Acceleration-Time Graph):
Slope has no common kinematic interpretation.
Area under the curve of a–t (between t1 and t2) gives the change in velocity, \Delta v = \int{t1}^{t_2} a(t)\,dt. This is particularly useful for finding velocity when acceleration is not constant but given as a function of time.
Velocity–position–acceleration relationships (three common cases)
These relationships are fundamental for solving problems where acceleration is not constant.
1) a is a function of time: a(t). To find velocity and position, integrate sequentially.Velocity: v(t) = v0 + \int0^t a(\tau)\,d\tau (where v_0 is the initial velocity at t=0).
Position: x(t) = x0 + \int0^t v(\tau)\,d\tau (where x_0 is the initial position at t=0).
2) a is a function of position: a(x). Use the chain rule to relate acceleration, velocity, and position directly.
Derivation: a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}
This leads to: v\,dv = a(x)\,dx
Integrating both sides: \int{v0}^v v'\,dv' = \int{x0}^x a(x')\,dx' \Rightarrow \frac{v^2}{2} - \frac{v0^2}{2} = \int{x_0}^x a(x')\,dx'
This allows you to solve for v as a function of x. Then, you can use dx/dt = v(x) to find x(t) by integrating.
3) a is a function of velocity: a = a(v). This case often involves differential equations.
From the definition of acceleration: \frac{dv}{dt} = a(v)
Rearranging to separate variables: dt = \frac{dv}{a(v)}
Integrating to find time as a function of velocity: t = \int{v0}^v \frac{dv'}{a(v')}. This allows you to find v(t) by inverting the function.
Once v(t) is known, position can be found by integrating velocity: x(t) = x0 + \int0^t v(\tau)\,d\tau.
Important derived formula (example of converting v(x) to a(x))
Given a velocity–position relation v = v(x), the acceleration is obtained using the chain rule:
a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}
This formula is extremely useful when the velocity of an object is given directly as a function of its position, and you need to find its acceleration. It bypasses the need to find time explicitly if not required.
Example: If v(x) = -2x + 4, then first find \frac{dv}{dx} = -2. Then, substitute into the formula:
a = v \frac{dv}{dx} = (-2x+4)(-2) = 4x - 8.
This result shows that the acceleration is also a function of position in this specific case.
Graph interpretation and conversions (study tips)
x–t graph (Displacement-Time Graph)
Slope = v(t); the sign of the slope indicates the direction of motion (positive slope = positive direction, negative slope = negative direction, zero slope = at rest).
If x–t is linear (straight line), velocity is constant (zero acceleration).
If x–t is curved, velocity is changing, implying acceleration is nonzero. A parabolic x-t graph indicates constant acceleration.
The point of maximum or minimum displacement (turning point) corresponds to where the velocity is zero.
v–t graph (Velocity-Time Graph)
Slope = a(t); a constant slope implies constant acceleration. A horizontal line means constant velocity (zero acceleration).
Area under v–t curve between t1 and t2 equals displacement during that interval. Area above the t-axis is positive displacement; area below is negative displacement.
The total distance traveled is the sum of the absolute values of the areas.
a–t graph (Acceleration-Time Graph)
Area under a–t curve between t1 and t2 equals the change in velocity: \Delta v = \int{t1}^{t_2} a(t)\,dt.
This means if you have v0 as initial velocity, then v(t) = v0 + \text{Area under a-t graph from 0 to t}. A horizontal line indicates constant acceleration (e.g., free fall). A curved line indicates changing acceleration.
Converting between graphs (Systematic Approach)
From x–t to v–t: Take the slope of the x–t graph at each point in time. If x-t is linear, v-t will be a horizontal line. If x-t is parabolic, v-t will be a linear line.
From v–t to a–t: Take the slope of the v–t graph at each point in time. If v-t is linear, a-t will be a horizontal line. If v-t is curved, a-t will be a curved line.
From a–t to v–t: Integrate a(t) to get v(t). This means finding the area under the a-t graph, adding the initial velocity v_0 (change in velocity = area). The shape of the v-t graph will be one degree higher than a-t (e.g., constant a-t -> linear v-t).
From v–t to x–t: Integrate v(t) to get x(t). This means finding the area under the v-t graph, adding the initial position x_0 (displacement = area). The shape of the x-t graph will be one degree higher than v-t (e.g., linear v-t -> parabolic x-t).
Key relationships used in problems
Constant acceleration: These are the cornerstone of kinematics problems when acceleration is uniform. They assume motion in a straight line with constant acceleration.
v = u + a t (Velocity-time relation)
x = ut + \tfrac{1}{2} a t^2 (Displacement-time relation)
v^2 = u^2 + 2 a x (Velocity-displacement relation, useful when time is not given or required)
x = \tfrac{1}{2} (u + v) t (Alternative displacement-time relation)
Energy/kinetic energy relationships (when applicable): These can be extremely useful in problems involving conservative forces like gravity where mechanical energy is conserved. For example, in free fall, mgh + \frac{1}{2}mu^2 = constant (neglecting air resistance), which often simplifies calculations compared to kinematic equations, especially when dealing with heights and speeds at different points.
Selected worked examples (based on transcript content)
Example 1: v(x) and a(x) relationship
Given: velocity varies with position as a linear function: v(x) = -2x + 4. This implies that as position x changes, the velocity changes linearly.
Find acceleration as a function of x.
Solution:
First, calculate the derivative of v with respect to x: \frac{dv}{dx} = \frac{d}{dx}(-2x + 4) = -2.
Now, use the chain-rule relation for acceleration: a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}
Substitute the given v(x) and the calculated \frac{dv}{dx} into the formula:
a = (-2x+4)(-2) = 4x - 8.
Answer: a(x) = 4x - 8.
Significance: This example vividly demonstrates the application of the chain-rule technique to convert from a velocity-position function (v(x)) to an acceleration-position function (a(x)) when acceleration is not directly given as a function of time. Note that since a(x) depends on x, the acceleration is not constant.
Example 2: Rocket launched upward with piecewise acceleration (Lecture 08, Page 7)
Problem setup: A rocket is launched vertically upward. During its powered phase (where fuel is burning), it experiences a net upward acceleration of +4\text{ m/s}^2. Its initial velocity is 0. After 5 seconds, the fuel runs out, and the rocket then moves under the influence of gravity alone, with g = 10\text{ m/s}^2 downward. The highest point of the rocket's trajectory occurs when its velocity becomes zero. After reaching its peak, it accelerates downward at g until it returns to the ground.
Calculations:
Powered phase (0 to 5 s): During this phase, acceleration is constant (a = +4\text{ m/s}^2).
Velocity at the end of the powered phase (t=5\text{ s}): v1 = v0 + a t = 0 + 4(5) = 20\text{ m/s}. This is the maximum velocity achieved by the rocket during its powered flight.
Displacement during the powered phase (s1): s1 = v_0 t + \tfrac{1}{2} a t^2 = 0(5) + 0.5 (4)(5)^2 = 0 + 0.5(4)(25) = 50\text{ m}. The rocket reaches a height of 50 m with an upward velocity of 20 m/s when the fuel runs out.
Coast phase (5 to 7 s): After 5 seconds, the fuel is exhausted, and the rocket is only under gravity. Acceleration is now a = -g = -10\text{ m/s}^2 (negative because it's downward).
The initial velocity for this phase is v_{initial} = 20\text{ m/s}. The rocket continues to move upward, decelerating until its velocity becomes zero at the peak.
Time until velocity reaches zero (tc): Using v = u + at, we have 0 = 20 + (-10) tc \Rightarrow t_c = \frac{20}{10} = 2\text{ s}. So, velocity becomes zero at 5 + 2 = 7\text{ s}.
Displacement during coast (s_2): Using s = ut + \tfrac{1}{2} a t^2 with u=20\text{ m/s} and t=2\text{ s}:
s_2 = 20(2) - 0.5(10)(2)^2 = 40 - 0.5(10)(4) = 40 - 20 = 20\text{ m}. The rocket rises an additional 20 m during this phase.
Peak height above launch: The maximum height (h_{max}) is the sum of displacements from both phases.
h{max} = s1 + s_2 = 50 + 20 = 70\text{ m}.
Fall from peak to ground: The rocket starts falling from rest (u=0) at the peak height (h = 70\text{ m}) under gravity (g = 10\text{ m/s}^2).
Time to fall (t_f): Using x = ut + \tfrac{1}{2} a t^2, with x=70\text{ m}, u=0, a=10\text{ m/s}^2 (if measuring displacement downward from peak):
70 = 0 + \tfrac{1}{2}(10) tf^2 \Rightarrow 70 = 5 tf^2 \Rightarrow tf^2 = 14 \Rightarrow tf = \sqrt{14} \approx 3.74\text{ s}.
Total journey time: The total time from launch until it hits the ground is the sum of times for all three phases.
t_{total} = \text{powered phase time} + \text{coast phase time} + \text{fall phase time}
t_{total} = 5\text{ s} + 2\text{ s} + \sqrt{14}\text{ s} \approx 10.74\text{ s}.
Graph expectations:
v–t graph: Will be piecewise linear. Initially, a linear increase from 0 to 20 m/s (slope +4 m/s²). Then, a linear decrease from 20 m/s to 0 m/s (slope -10 m/s²), crossing the t-axis at 7 s. Finally, a continued linear decrease into negative velocities (slope -10 m/s²) as it falls back to the ground.
x–t graph: Will be a curve that rises parabolically during the powered phase, then continues to rise (but less steeply) and curves downward during the coast phase, reaching a maximum height at t \approx 7\text{ s}. After that, it curves more steeply downward as it accelerates back towards the ground, eventually reaching x=0.
Significance: This example provides a complete analysis of motion that involves multiple phases with different constant accelerations, demonstrating how to combine kinematic equations for sequential segments of motion to find total time, displacement, and peak height. It also highlights graphical interpretations in such complex scenarios.
Example 3: A drunkard’s walk in a straight line (NCERT-style) – 5 forwards, 3 backwards, step length 1 m and duration 1 s per step
Pattern: This is a discrete motion problem. Each full cycle of the drunkard's movement consists of 5 steps forward (each +1 m) followed by 3 steps backward (each -1 m). Each step takes 1 second. Therefore, the duration of one full cycle is 5\text{ s} + 3\text{ s} = 8\text{ s}. The net progress (displacement) per cycle is 5\text{ m} - 3\text{ m} = +2\text{ m}.
Task: Plot the x–t graph and determine the time it takes for the drunkard to fall into a pit located 9 m away from the starting point.
Approach:
After k full cycles, the total distance covered is 2k meters (net displacement), and the total time elapsed is 8k seconds.
We need to reach 9 m. Let's analyze cycle by cycle:
After 1 full cycle (t=8\text{ s}), displacement = 2 m.
After 2 full cycles (t=16\text{ s}), displacement = 4 m.
Now, the drunkard needs to cover an additional 9\text{ m} - 4\text{ m} = 5\text{ m}. This will happen in the forward-stepping part of the next cycle (the 3rd cycle).
Starting the 3rd cycle from x=4\text{ m} at t=16\text{ s}, the drunkard takes 5 forward steps, each taking 1 s and moving 1 m forward.
The pit is at 9 m. The drunkard will reach 9 m by taking exactly 5 forward steps in this cycle (from 4 m to 9 m).
Time taken for these 5 forward steps = 5 \times 1\text{ s} = 5\text{ s}.
The total time to reach 9 m is the time for 2 full cycles plus the time for these 5 forward steps: t = 16\text{ s} + 5\text{ s} = 21\text{ s}.
Answer: It takes 21 seconds to reach 9 meters. The pit is reached during the forward phase of the 3rd cycle. The earliest time to hit 9 m under the given pattern is 21 s, as any backward steps during this cycle would take them away from the pit.
Conceptual takeaway: Problems involving discrete-step motion can be effectively analyzed by calculating the net progress per cycle and then considering the individual steps within the final partial cycle. The x–t graph will typically have stair-step segments, with periods of increasing displacement during forward motion and decreasing displacement during backward motion.
Example 4: Height 500 m problem with two particles (Lecture 08, Page 12–13)
Setup: Two particles, A and B, are considered with motion under gravity from a height of 500 m. We use g = 10\text{ m/s}^2.
Particle A is thrown upward from height 500 m with initial velocity v{A0} = 75\text{ m/s}. Its acceleration is aA = -g = -10\text{ m/s}^2 (negative because gravity acts downward while initial velocity is upward).
Particle B is released from rest from the same height 500 m, so its initial velocity is v{B0} = 0\text{ m/s}. Its acceleration is aB = -g = -10\text{ m/s}^2.
Solutions:
Calculations for Particle A: Vertical motion under gravity with initial upward velocity.
Velocity as a function of time: vA(t) = v{A0} + a_A t = 75 - 10 t.
Displacement from the ground as a function of time: Since it starts at 500 m, xA(t) = x0 + v{A0} t + \tfrac{1}{2} aA t^2 = 500 + 75 t - 5 t^2.
Peak height: Reached when the velocity becomes zero, i.e., v_A(t) = 0.
0 = 75 - 10 t{rise} \Rightarrow t{rise} = \frac{75}{10} = 7.5\text{ s}. This is the time to reach the highest point from launch.
Extra height reached during the rise (from 500 m): Using v^2 = u^2 + 2ax, with v=0, u=75, a=-10, and x=\Delta h. (0)^2 = (75)^2 + 2(-10) \Delta h \Rightarrow 0 = 5625 - 20 \Delta h \Rightarrow \Delta h = \frac{5625}{20} = 281.25\text{ m}.
Peak height above ground: h_{peak} = 500\text{ m} + 281.25\text{ m} = 781.25\text{ m}.
Time to fall from peak to ground: The particle falls 781.25 m from rest from its highest point. Using h = \tfrac{1}{2} g t^2 (setting u=0 and taking downward as positive for fall):
781.25 = \tfrac{1}{2}(10) t{fall}^2 \Rightarrow 781.25 = 5 t{fall}^2 \Rightarrow t{fall}^2 = 156.25 \Rightarrow t{fall} = \sqrt{156.25} = 12.5\text{ s}.
Total time for Particle A to reach the ground: t{A,ground} = t{rise} + t_{fall} = 7.5\text{ s} + 12.5\text{ s} = 20.0\text{ s}.
Calculations for Particle B: Free fall from rest.
Particle B is simply dropped from 500 m (u=0) with acceleration a=-10\text{ m/s}^2.
Time to ground (t_B): Using x = ut + \tfrac{1}{2} a t^2, where x=-500\text{ m} (final position relative to initial) or h=500\text{ m} (distance fallen):
500 = 0 + \tfrac{1}{2}(10) tB^2 \Rightarrow 500 = 5 tB^2 \Rightarrow tB^2 = 100 \Rightarrow tB = \sqrt{100} = 10.0\text{ s}.
Graph expectations for both particles:
a–t graph: For both particles, the acceleration is constant and equal to -g. Thus, the a–t graph will be a horizontal line at -10 m/s² for all times significant to their motion. (Particle A initially has an upward velocity but its acceleration due to gravity is always -g).
v–t graph: For Particle A, the v-t graph starts at +75 m/s, linearly decreases with a slope of -10 m/s², crosses the t-axis at 7.5 s (peak height), and then continues linearly into negative values as it falls. For Particle B, the v-t graph starts at 0 m/s and linearly decreases with a slope of -10 m/s², continuously becoming more negative as it falls (straight line with negative slope starting from the origin).
x–t graph: For Particle A, the x-t graph starts at 500 m, curves upward to a peak at 781.25 m (at 7.5 s), and then curves downward parabolically until it reaches 0 m (ground) at 20.0 s. For Particle B, the x-t graph starts at 500 m and smoothly curves downward parabolically, reaching 0 m (ground) at 10.0 s. It shows monotonic decrease in height without any turning point.
Example 5: Height of 500 m; A and B graphs (displacement–time, velocity–time, and acceleration–time) – conceptual results
Particle A (thrown upward): Its velocity is initially positive (+75 m/s) and decreases linearly due to constant negative acceleration. It becomes zero at 7.5 s (peak) and subsequently becomes increasingly negative as it falls. Its displacement (x-t) graph is an inverted parabola starting from 500m, rising to a peak, and then falling back to 0m. The acceleration graph is a constant negative horizontal line.
Particle B (released from rest): Its initial velocity is zero and it immediately starts accelerating downwards, so its velocity becomes increasingly negative linearly. Its displacement (x-t) graph is a simple downward curving parabola from 500m to 0m. The acceleration graph is also a constant negative horizontal line, identical to particle A's acceleration graph after particle A starts falling freely.
Acceleration for both: Both particles experience the same constant acceleration a = -g. Therefore, their a-t graphs are identical (a horizontal line at -10 m/s²).
Displacements: A's x(t) equation xA(t) = 500 + 75 t - 5 t^2 reflects its initial upward motion and subsequent downward motion. B's x(t) equation xB(t) = 500 - 5 t^2 simply describes its continuous fall from 500 m. The x-t graphs visually represent these unique trajectories.
Example 6: Velocity vs displacement graph problem (displacement–time and velocity–time relationships)
Given a velocity–displacement relation (graph schematic, e.g., v(x) is shown). The typical question asks to choose the corresponding acceleration–displacement relation, or perhaps a different graph form.
Conceptual approach:
The key formula to use is a = v \frac{dv}{dx}. This means we need the value of velocity at a given position and the slope of the v-x graph at that same position.
If velocity increases with displacement (v-x graph has a positive slope), and velocity v is positive, then acceleration is positive.
If velocity decreases with displacement (v-x graph has a negative slope), and velocity v is positive, then acceleration is negative (slowing down).
If velocity increases with displacement (v-x graph has a positive slope), and velocity v is negative, then acceleration is negative (speeding up in the negative direction, e.g., if v = -5 m/s and dv/dx = 1, then a = -5, meaning increasingly negative velocity).
If velocity decreases with displacement (v-x graph has a negative slope), and velocity v is negative, then acceleration is positive (slowing down in the negative direction, e.g., if v = -5 m/s and dv/dx = -1, then a = +5, meaning velocity is becoming less negative).
To determine the exact functional form of a(x), one must derive the equation for the v-x curve, find its derivative dv/dx, and then multiply it by v(x). Analyzing the slope and the sign of v can quickly narrow down the options for a(x) graph.
Example 7: Free-fall and energy-based problems (ball dropped from height, bounce height, etc.)
These problems often involve constant gravitational acceleration (g) and can frequently be simplified using energy conservation principles, assuming negligible air resistance.
Typical kinematic relations for vertical motion under gravity:
Velocity at height h (or after displacement \Delta x): v^2 = v_0^2 + 2a (\Delta x). For free fall, a = -g. If falling downward, and taking downward as positive, v^2 = u^2 + 2g h.
For vertical throw with initial velocity u upward and acceleration a = -g (taking upward as positive):
Height as a function of time: y(t) = h_0 + u t - \tfrac{1}{2} g t^2
Velocity as a function of time: v(t) = u - g t
Energy conservation approach: When only gravity (a conservative force) is doing work, mechanical energy (Kinetic Energy + Potential Energy) is conserved.
KEi + PEi = KEf + PEf
\tfrac{1}{2} m vi^2 + m g hi = \tfrac{1}{2} m vf^2 + m g hf
This allows direct relation between velocities and heights without explicitly solving for time, which can be very efficient, especially for bounce problems or finding maximum heights and impact velocities.
Practice problems (concepts and quick solutions)
Problem: A particle moving along x with velocity given by V = V0 − Kx
Determine the acceleration–time (a–t) plot during the interval as x increases from 0 to some value.
Approach: This is a v(x) problem. Use the relation a = v \frac{dv}{dx}. First, differentiate V = V0 - Kx with respect to x to find \frac{dV}{dx} = -K. Then, a = (V0 - Kx)(-K) = -KV0 + K^2x. So, acceleration is a linear function of x. To get the a-t plot, you then need to find x(t) by integrating dx/dt = V0 - Kx, and substitute x(t) into the a(x) expression. This results in an exponential dependence for a(t), since x(t) will be exponential.
Result sketch: Acceleration depends on x linearly; the a–t plot is determined by how x evolves with time via v. Since V decreases linearly with x, a is a linear function of x. The tricky part is translating this to a vs. t, which requires solving for x(t) and then a(t).
Problem: A train moving along a track with a speed–time profile given; compute maximum acceleration and distance between certain time marks
Approach: For maximum acceleration, identify the segment of the speed–time graph (v-t) with the steepest slope (largest magnitude). The slope of the v-t graph represents acceleration. To compute displacement (or distance), calculate the area under the v-t curve for the specified time segments. Remember that area above the t-axis is positive displacement, below is negative. For total distance, sum the absolute values of the areas.
Problem: A body moving with a given a–t graph; determine speed at certain times t = 2 s and t = 3 s
Approach: To find velocity from an a-t graph, integrate a(t), which means finding the area under the a–t curve. Start with the initial velocity v0 (if given). The change in velocity over an interval is the area under the a-t graph for that interval (\Delta v = \text{Area}{a-t}). So, v(t) = v_0 + \text{Area from 0 to t}. Then, if required, integrate v(t) to obtain position.
Problem: Displacement and distance from a position–time graph
Distinguish between displacement (net change in position from start to end point) and distance traveled (total length of the path taken, always positive). From an x-t graph: displacement is simply xf - xi. For total distance, identify all segments where motion changes direction (where the slope of x-t changes sign or where v=0), calculate distance for each segment (absolute value of change in x), and sum them up.
Use area under velocity curves and/or slope of position curves to extract information. Remember, the slope of x-t gives velocity. If the velocity goes negative, the object is moving backward.
Quick reference: common equations you should memorize
Basic definitions
Speed: v_{\text{speed}} = \frac{\text{distance travelled}}{\text{time}}
Velocity: Instantaneous velocity, \mathbf{v} = \frac{d\mathbf{x}}{dt} . Average velocity, \mathbf{v}_{avg} = \frac{\Delta \mathbf{x}}{\Delta t}
Displacement: Final position minus initial position, \Delta \mathbf{x} = \mathbf{x}f - \mathbf{x}i
Acceleration: Instantaneous acceleration, \mathbf{a} = \frac{d\mathbf{v}}{dt} . Average acceleration, \mathbf{a}_{avg} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{\mathbf{v}-\mathbf{u}}{t}
Constant-acceleration (kinematic) equations
Valid only when acceleration is constant over the time interval.v = u + a t (Final velocity = Initial velocity + Acceleration × Time)
x = u t + \tfrac{1}{2} a t^2 (Displacement = Initial velocity × Time + Half × Acceleration × Time²)
v^2 = u^2 + 2 a x (Final velocity² = Initial velocity² + 2 × Acceleration × Displacement)
x = \tfrac{1}{2} (u + v) t (Displacement = Average velocity × Time)
Relationship between a, v, and x (case when a = a(x))
This chain rule is crucial when acceleration depends on position:
a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}
Integrating form (useful for finding velocity as a function of position): \int v\,dv = \int a(x)\,dx \Rightarrow \frac{v^2}{2} = \int a(x)\,dx + C (The constant C is determined by initial conditions, e.g., v0 at x0).
Velocity from a(v) (acceleration as a function of velocity)
This applies when acceleration depends on velocity (e.g., drag forces):
\frac{dv}{dt} = a(v) \Rightarrow dt = \frac{dv}{a(v)}
Integrating to find time as a function of velocity: t = \int{v0}^v \frac{dv'}{a(v')}
Once v(t) is determined (often by inverting the above integral), position can be found:
dx = v(t)\,dt \Rightarrow x(t) = x0 + \int0^t v(\tau)\,d\tau
Energy-style relation for vertical motion under gravity (often derived from v^2 = u^2 + 2ax)
For an object moving vertically under constant gravity, taking upward as positive and a = -g:
v^2 = u^2 - 2g \Delta y (where \Delta y is the vertical displacement).
This mirrors the conservation of mechanical energy principle (\tfrac{1}{2} m v^2 + mgy = \text{constant}), effectively relating changes in kinetic and potential energy.
Practical exam tips
Always identify which graph you are dealing with (x–t, v–t, or a–t) and what physical quantity you need to extract (distance, displacement, time, velocity, acceleration, etc.). Misinterpreting the graph type is a common error.
Use slope-intercept interpretations: Remember that the slope of an x–t graph gives instantaneous velocity, the slope of a v–t graph gives instantaneous acceleration, and the area under a v–t graph gives displacement. The area under an a-t graph gives the change in velocity.
Be precise with sign conventions for velocity and acceleration: Consistently define a positive direction (e.g., upward or rightward) and stick to it throughout the problem. A negative sign means motion or acceleration in the opposite direction.
For gravity problems, use the given value of g (often approximated to 10 m/s² for simplicity, but use 9.8 or 9.81 m/s² if specified). Always assign a direction to g (e.g., -10 m/s² if upward is positive).
In problems involving piecewise accelerations (like the rocket example), break the motion into distinct segments where acceleration is constant. Solve for the final velocity and position of one segment, which then become the initial conditions for the next segment. Stitch these results together to find total time, distance, and height.
When converting v(x) to a(x), rigorously apply the chain rule relation a = v \frac{dv}{dx}. Be careful with the differentiation and multiplication steps.
For projectile-like problems or those involving vertical motion, you can often use energy-like relations (v^2 = u^2 + 2ax or conservation of mechanical energy) to find peak heights or velocities at specific positions without needing to calculate time, which can save computation steps.
Pay attention to units consistently. All calculations should use SI units (meters, seconds, kilograms).
Understand average vs. instantaneous quantities: Distinguish between average values (over an interval) and instantaneous values (at a specific point in time). The concepts of slope and area are key to this distinction.
Summary takeaways
Motion in one dimension is precisely described by fundamental calculus relationships: velocity is the time derivative of position (v = dx/dt), acceleration is the time derivative of velocity (a = dv/dt and a = d^2x/dt^2), and the powerful chain-rule relation (a = v dv/dx) is essential when quantities depend on position.
Graphical analysis is a robust tool: slopes of cinematic graphs directly provide rates of change (velocity from x-t, acceleration from v-t), while areas beneath the curves yield cumulative quantities (displacement from v-t, change in velocity from a-t).
Effective problem-solving often depends on segmenting complex motion into intervals of constant acceleration. Applying the appropriate kinematic equations to each interval and then combining the results systematically leads to a complete solution for the entire motion.
These concepts have significant real-world relevance, underpinning various fields from trajectory planning in aerospace engineering (e.g., rocket launches), predicting fall times in construction or sports, to understanding how velocity and position evolve in diverse mechanical and physical systems. Mastery of these fundamentals is critical