**Motion** is the change in position of an object with respect to time. In this chapter, we study **rectilinear motion** (motion along a straight line), confining ourselves to one-dimensional kinematics without discussing causes of motion.
Key concepts:
**Instantaneous velocity** is the velocity of an object at a specific instant of time. While average velocity tells us motion over a time interval, instantaneous velocity tells us how fast an object is moving at that exact moment.
**Mathematical Definition:**
v = lim(Δt→0) [Δx/Δt] = dx/dt
Where:
**Graphical Interpretation:**
Given: x = 0.08t³
To find velocity at t = 4 s, calculate Δx/Δt for decreasing values of Δt:
As Δt → 0, the ratio approaches **3.84 m/s**, which equals dx/dt at t = 4 s.
Using calculus: dx/dt = d(0.08t³)/dt = 0.24t² = 0.24(16) = 3.84 m/s ✓
**Instantaneous speed** is the magnitude of instantaneous velocity:
Speed = |v|
**Key Distinction:**
**Problem:** Position of object: x = a + bt² where a = 8.5 m, b = 2.5 m/s². Find velocity at t = 0 s, t = 2.0 s, and average velocity between t = 2.0 s and t = 4.0 s.
**Solution:**
v = dx/dt = d(a + bt²)/dt = 2bt = 5.0t m/s
At t = 0 s: v = 0 m/s
At t = 2.0 s: v = 5.0 × 2.0 = 10 m/s
Average velocity = [x(4) – x(2)]/[4 – 2]
x(4) = 8.5 + 2.5(16) = 48.5 m
x(2) = 8.5 + 2.5(4) = 18.5 m
Average velocity = (48.5 – 18.5)/2 = **15 m/s**
**Average acceleration** is the change in velocity divided by the time interval during which the change occurs:
a = Δv/Δt = (v₂ – v₁)/(t₂ – t₁)
Where:
**Graphical Interpretation:** On a velocity-time (v-t) graph, average acceleration is the **slope of the straight line** connecting points (v₁, t₁) and (v₂, t₂).
**Instantaneous acceleration** is the acceleration at a specific instant:
a = lim(Δt→0) [Δv/Δt] = dv/dt
**Key Feature:** On a v-t graph, instantaneous acceleration is the **slope of the tangent** to the curve at that instant.
Since velocity has both magnitude and direction, acceleration can result from:
1. Change in speed (magnitude) only
2. Change in direction only
3. Changes in both speed and direction
Acceleration can be:
For motion with constant acceleration:
Specific cases:
1. Object moving in positive direction with positive acceleration: line has positive slope, above time axis
2. Object moving in positive direction with negative acceleration: line has negative slope, starts above time axis
3. Object moving in negative direction with negative acceleration: line has negative slope, below time axis
4. Object changes direction: v-t line crosses time axis at the instant of direction change
If velocity is v₀ at t = 0 and v at time t:
**v = v₀ + at** ... (2.4)
The **area under the v-t curve** between two time instants equals the **displacement** during that interval.
**Proof for constant velocity:** For object moving with constant velocity u, the v-t graph is a horizontal straight line. Area = height × base = u × T = uT = displacement over time T. ✓
**Why is area = displacement?**
Kinematic equations relate five quantities: displacement (x), time (t), initial velocity (v₀), final velocity (v), and acceleration (a).
Starting with: a = (v – v₀)/t
**v = v₀ + at** ... (2.4)
This is graphically represented by a straight line on a v-t graph.
From v-t graph with constant acceleration:
Displacement = Area under v-t curve = Area of trapezoid
Area = (1/2)(v₀ + v) × t
Therefore: **x = [(v₀ + v)/2] × t** ... (2.5)
Substituting v = v₀ + at:
x = [(v₀ + v₀ + at)/2] × t = [(2v₀ + at)/2] × t
**x = v₀t + (1/2)at²** ... (2.6)
**Alternative form:**
Since average velocity = (v₀ + v)/2 for constant acceleration:
**x = v̄ × t** where v̄ = (v₀ + v)/2 ... (2.7)
From equation (2.4): t = (v – v₀)/a
Substitute into equation (2.5):
x = [(v₀ + v)/2] × [(v – v₀)/a]
x = [(v₀ + v)(v – v₀)]/[2a]
x = (v² – v₀²)/(2a)
**v² = v₀² + 2ax** ... (2.8)
This equation is useful when time is not known.
For motion with constant acceleration, assuming initial position x₀ = 0 at t = 0:
1. **v = v₀ + at**
2. **x = v₀t + (1/2)at²**
3. **v² = v₀² + 2ax**
**General forms** (when initial position is x₀):
1. **v = v₀ + at**
2. **x = x₀ + v₀t + (1/2)at²**
3. **v² = v₀² + 2a(x – x₀)**
**Important Notes:**
By definition: a = dv/dt
dv = a dt
Integrating: ∫dv = ∫a dt
v = v₀ + at ✓
Since v = dx/dt:
dx = v dt = (v₀ + at) dt
Integrating: ∫dx = ∫(v₀ + at) dt
x = v₀t + (1/2)at² + C
At t = 0, x = x₀, so C = x₀
x = x₀ + v₀t + (1/2)at² ✓
Also, a = v(dv/dx), so v dv = a dx
Integrating: ∫v dv = ∫a dx
(v² – v₀²)/2 = a(x – x₀)
v² = v₀² + 2a(x – x₀) ✓
**Problem:** Ball thrown upward with v₀ = 20 m/s from top of building 25 m high. Find: (a) maximum height above launch point, (b) time to hit ground. Take g = 10 m/s².
**Solution:**
**(a) Maximum Height:**
At maximum height, v = 0
Using: v² = v₀² + 2a(y – y₀)
0 = (20)² + 2(–10)(y – y₀)
0 = 400 – 20(y – y₀)
**y – y₀ = 20 m**
**(b) Time to Hit Ground:**
**Method 1 (Split Motion):**
0 = 45 + 0 – (1/2)(10)t₂²
t₂² = 9; **t₂ = 3 s**
**Method 2 (Direct Approach):**
Using: y = y₀ + v₀t + (1/2)at²
0 = 25 + 20t – 5t²
5t² – 20t – 25 = 0
t² – 4t – 5 = 0
(t – 5)(t + 1) = 0
**t = 5 s** (rejecting negative time)
Method 2 is preferred as we need not analyze the path separately.
**Definition:** Free fall is motion under gravity alone when air resistance is neglected.
For motion near Earth's surface with height << Earth's radius, g = constant = 9.8 m/s²
**For object released from rest** (v₀ = 0) **at y = 0, moving downward:**
Taking downward as negative direction:
**Graphs for free fall:**
1. a-t graph: horizontal line at a = –g (constant)
2. v-t graph: straight line with negative slope (linear decrease from 0)
3. y-t graph: parabola opening downward (quadratic dependence)
**Statement:** Distances traveled by an object falling from rest during successive equal time intervals are in ratio 1:3:5:7:9:... (odd numbers).
**Proof:**
Using y = –(1/2)gt² with initial velocity = 0
After intervals 0, τ, 2τ, 3τ, ...:
Distances in successive intervals:
Ratio = (1/2)gτ² : (3/2)gτ² : (5/2)gτ² = **1:3:5** ✓
**Problem:** Derive expression for stopping distance when brakes cause deceleration a.
**Solution:**
When brakes are applied: final velocity v = 0, initial velocity = v₀
Using: v² = v₀² + 2a × s
0 = v₀² + 2(–a)s (deceleration = –a)
**s = v₀²/(2a)**
**Key Point:** Stopping distance is proportional to the **square of initial velocity** (s ∝ v₀²)
Example: If initial velocity doubles, stopping distance increases by factor of 4.
This is why speed limits are strictly enforced in populated areas — the non-linear relationship between velocity and stopping distance makes high-speed collisions far more severe.
**Relative velocity** of object A with respect to object B is the velocity of A as observed from the reference frame of B.
In one dimension:
**v_{AB} = v_A – v_B**
Where:
v_{BA} = –v_{AB}
(If A moves at +5 m/s relative to B, then B moves at –5 m/s relative to A)
**Case 1: Objects moving in same direction**
**Case 2: Objects moving in opposite directions**
**Case 3: Object moving perpendicular to observer's motion**
Relative velocity is crucial for:
---
| Equation | When to Use | Limitation |
|----------|------------|-----------|
| v = v₀ + at | When time and velocities given | Constant a only |
| x = v₀t + (1/2)at² | When initial velocity, time, acceleration known | Constant a only |
| v² = v₀² + 2ax | When time not given | Constant a only |
| x = [(v₀ + v)/2]t | Direct with average velocity | Constant a only |
| a = dv/dt | Instantaneous acceleration | Works for any a(t) |
| v = dx/dt | Instantaneous velocity | Works for any x(t) |
1. **Sign Convention:** Choose consistently; typically upward/rightward = positive
2. **Instantaneous vs. Average:** Instantaneous values are derivatives; average values are changes divided by time intervals
3. **Graphical Method:**
4. **Kinematic Equations:** Valid ONLY for constant acceleration
5. **Free Fall:** g = 9.8 m/s² ≈ 10 m/s²; always acts downward (negative in upward convention)
6. **Relative Motion:** Always subtract reference velocities correctly
7. **Calculus Approach:** Integration gives displacement from velocity; differentiation gives velocity from position
Q1. The position of a particle is given by x = 5 + 3t. What is its velocity?
Answer: A — Velocity v = dx/dt = d(5 + 3t)/dt = 3 m s⁻¹ (constant).
Q2. From a position-time graph, how is instantaneous velocity at a point determined?
Answer: B — Instantaneous velocity is v = dx/dt, which is the slope of the tangent to the x-t curve at that instant.
Q3. A car travels 100 m in 5 s, then 50 m in 3 s. What is its average speed?
Answer: A — Average speed = total distance/total time = (100 + 50)/(5 + 3) = 150/8 = 18.75 m s⁻¹.
Q4. For x = 0.08t³, what is the velocity at t = 4.0 s?
Answer: C — v = dx/dt = 0.24t²; at t = 4.0 s, v = 0.24 × 16 = 3.84 m s⁻¹ (matches Table 2.1 limit).
Q5. Which statement about instantaneous speed and instantaneous velocity is correct?
Answer: B — Speed is a scalar (magnitude only) equal to |velocity|; velocity is a vector with direction, so speed = |velocity| at any instant.
Q6. If a velocity-time graph shows a straight line with positive slope, what does this indicate?
Answer: B — The slope of a v-t graph equals acceleration; a positive slope means acceleration is constant and positive.
Q7. Average velocity differs from average speed. Which is NOT a correct statement?
Answer: C — For circular motion, displacement = 0 (returns to start), so average velocity = 0, but average speed > 0; magnitudes are never equal.
Q8. A particle's position is x = 10t² + 5t + 2. What is its acceleration at t = 3 s?
Answer: A — v = dx/dt = 20t + 5; a = dv/dt = 20 m s⁻² (constant for all t, independent of t = 3 s).
Q9. In the limiting process to find instantaneous velocity, as Δt approaches zero, the secant line connecting two points on the position-time curve approaches:
Answer: C — The secant line's slope = Δx/Δt; as Δt → 0, the secant becomes the tangent, whose slope is dx/dt = instantaneous velocity.
Q10. Why did Galileo define acceleration as rate of change of velocity with *time* rather than with *distance*?
Answer: B — Galileo's free fall experiments showed acceleration is constant with time but varies with distance, making time the physically meaningful variable.
What is the mathematical definition of instantaneous velocity?
Instantaneous velocity v = lim(Δt→0) Δx/Δt = dx/dt, the slope of the position-time graph at that instant.
How do you find velocity at a specific time from a position-time graph?
Draw the tangent line to the curve at that point; the slope of the tangent equals the instantaneous velocity.
What is the difference between instantaneous speed and instantaneous velocity?
Speed is the magnitude of velocity (always positive); velocity is a vector with both magnitude and direction (can be positive or negative).
Define average acceleration and write its formula.
Average acceleration a = (v₂ − v₁)/(t₂ − t₁) = Δv/Δt, the change in velocity divided by the time interval.
If x = 8.5 + 2.5t², what is the velocity at t = 2.0 s?
v = dx/dt = 5.0t, so at t = 2.0 s, v = 10 m s⁻¹.
What does the slope of a velocity-time graph represent?
The slope of a v-t graph equals the acceleration at that instant.
Why is acceleration defined as rate of change of velocity with time, not with distance?
Galileo proved through experiments on free fall that acceleration is constant with time but decreases with distance, making time the correct independent variable.
What is the SI unit of acceleration?
The SI unit of acceleration is m s⁻² (metres per second squared).
In the limiting process Δt → 0, what does the secant line become?
The secant line connecting two points on a curve becomes the tangent line at that point.
Why can objects be treated as point objects in kinematics?
Objects can be treated as point objects when their size is much smaller than the distance they travel in a reasonable time, making size negligible.
Define instantaneous velocity. Why is it different from average velocity? [2 marks]
Instantaneous velocity is v = lim(Δt→0) Δx/Δt = dx/dt (velocity at one specific instant). Average velocity is total displacement over total time interval. Instantaneous velocity can vary during motion; average velocity depends only on start and end points.
The position of an object is given by x = 2t³ − 5t + 10, where x is in metres and t is in seconds. Find (a) velocity at t = 2 s, (b) acceleration at t = 2 s, and (c) average velocity between t = 1 s and t = 3 s. Show all working steps. [5 marks]
Use v = dx/dt = 6t² − 5 for velocity; a = dv/dt = 12t for acceleration. For average velocity, calculate x(3) and x(1), then use Δx/Δt. Verify that instantaneous velocity ≠ average velocity unless motion is uniform.
Explain with a detailed derivation why instantaneous velocity is found as the slope of the tangent to a position-time graph, not the secant line. Include a graphical explanation of the limiting process Δt → 0. [6 marks]
Start with definition of average velocity as slope of secant (rise/run = Δx/Δt). Show that as Δt decreases, the two points get closer and secant approaches tangent. Use Table 2.1 concept (Δx/Δt → constant value) to show limit equals dx/dt. Draw or describe a graph showing how secant becomes tangent. Explain why dx/dt is the instantaneous rate of change.
Practice with interactive flashcards, mind maps, upload your own chapters and get AI study kits instantly
Try StudyOS Free →