Physics — Classical Mechanics — Kinematics — One-Dimensional Motion
Physics — Classical Mechanics — Kinematics — One-Dimensional Motion. Practice questions to deepen understanding of one-dimensional motion in classical mechanics kinematics. Online physics practice with full solutions and step-by-step explanations.
1D physics kinematics practice — 50 questions: velocity, acceleration, motion equations, x-t and v-t graphs, free fall. Classical mechanics for beginners.
Kinematics — motion in one dimension.
Question 1
2.00 pts
📚 Basic definition:
What is kinematics?
Explanation:
💡 Detailed explanation:
Kinematics - description of motion! 📚
| 🎯 What is kinematics? Kinematics: The branch of physics dealing with the description of motion Answers questions: • Where? Position (x) • How fast? Velocity (v) • How does the velocity change? Acceleration (a) Without asking: "Why does the body move?" 🔍 The important difference:
💡 Examples of kinematic questions: • Where is the vehicle in 5 seconds? • What's the speed of the ball when it hits the ground? • How long until the runner reaches the finish? |
Question 2
2.00 pts
📍 Concepts:
What is the difference between position and distance?
Explanation:
💡 Detailed explanation:
Position VS distance! 📍
| 📐 The definitions: Position - x: • Vector - has direction! • Measured relative to origin (0) • Can be positive (+) or negative (-) • Units: meter (m) Example: x = +5m (5 meters right of origin) x = -3m (3 meters left of origin) Distance - d: • Scalar - no direction • Length of the actual path traveled • Always positive (≥ 0) • Units: meter (m) Example: Going from x=0 to x=5 then back to x=2: distance = 5 + 3 = 8m position change = 2m 💡 Key insight: If you walk in a circle and return: • Distance = circumference (large number) • Position change = 0 (back to start!) |
Question 3
2.00 pts
📏 Displacement:
A body moves from x₁=3m to x₂=8m
What is the displacement?
Explanation:
💡 Detailed explanation:
Displacement! 📏
| 📐 Displacement definition: Formula: Δx = x₂ - x₁ Δx = displacement x₂ = final position x₁ = initial position In our example: x₁ = 3m (initial position) x₂ = 8m (final position) Δx = 8m - 3m = 5m The displacement is positive = the body moves right 💡 Displacement is a vector: • Positive (+) = motion right/up • Negative (-) = motion left/down • Zero (0) = the body returned to its initial position ⚠️ Note: Displacement ≠ distance traveled Displacement only depends on start and end points, not on the actual path! |
Question 4
2.00 pts
🚗 Average velocity:
What is the formula for average velocity?
Explanation:
💡 Detailed explanation:
Average velocity! 🚗
| 📚 The definition: Average velocity: v̄ = Δx/Δt v̄ = average velocity Δx = displacement Δt = time elapsed Units: m/s (meters per second) 🔍 Important differences:
💡 Example: Round-trip 10 km in each direction, total time 1 hour: • Distance = 20 km, average speed = 20 km/h • Displacement = 0, average velocity = 0! Big difference! |
Question 5
2.00 pts
⚖️ Unit conversion:
100 km/h = ? m/s
Explanation:
💡 Detailed explanation:
Unit conversion! ⚖️
| 📐 Conversion: From km/h to m/s: Divide by 3.6 or multiply by 1000/3600 Calculation: 100 km/h = 100 × (1000m) / (3600s) = 100,000m / 3600s = 27.777... m/s ≈ 27.8 m/s 💡 Quick way: 100 ÷ 3.6 = 27.8 Remember: • km/h → m/s: ÷ 3.6 • m/s → km/h: × 3.6 |
Question 6
2.00 pts
⚡ Instantaneous velocity:
What is instantaneous velocity?
Explanation:
💡 Detailed explanation:
Instantaneous velocity! ⚡
| 📚 The definition: Instantaneous velocity: The velocity at a specific moment v = lim(Δt→0) Δx/Δt or in calculus: v = dx/dt 🔍 The difference:
💡 Speedometer example: The car speedometer shows the instantaneous velocity at every moment. Even though your average speed for the trip might be 60 km/h, at a specific moment you might be going 80 or 40. |
Question 7
2.00 pts
🚀 Acceleration:
What is acceleration?
Explanation:
💡 Detailed explanation:
Acceleration! 🚀
| 📚 The definition: Acceleration: The rate of change of velocity a = Δv/Δt or: a = (v - v₀)/t v = final velocity v₀ = initial velocity t = time Units: m/s² (meters per second squared) 🔍 Meaning: Acceleration means the velocity is changing! Three cases: • a > 0: velocity increases (speeds up) • a < 0: velocity decreases (slows down) • a = 0: velocity is constant 💡 Example: From 0 to 30 m/s in 6s: a = (30-0)/6 = 5 m/s² Means: every second, the velocity grows by 5 m/s |
Question 8
2.00 pts
🔄 Sign of acceleration:
A vehicle moves to the right (v>0) and brakes.
What is the sign of the acceleration?
Explanation:
💡 Detailed explanation:
Sign of acceleration! 🔄
| 📐 Analysis: Situation: • Vehicle moves right: v > 0 (positive) • Brakes = slows = velocity decreases Numerical example: Time t₁: v₁ = +20 m/s Time t₂: v₂ = +10 m/s Δv = v₂ - v₁ = 10 - 20 = -10 m/s a = Δv/Δt = -10/Δt → negative! Conclusion: When braking (slowing): → velocity decreases → Δv negative → a negative 💡 Important note: Negative acceleration ≠ "moving in reverse"! It just means: opposing the current direction of motion. Possible cases: • Moving right, slowing → a < 0 • Moving left, slowing → a > 0 • Moving right, speeding up → a > 0 |
Question 9
2.00 pts
📏 Uniform motion:
What is uniform motion?
Explanation:
💡 Detailed explanation:
Uniform motion! 📏
| 📚 The definition: Uniform motion: Motion where the velocity is constant v = constant a = 0 📐 Properties: ✓ Constant velocity - doesn't change ✓ Zero acceleration - no change in velocity ✓ v-t graph: horizontal (straight) line ✓ x-t graph: straight tilted line 📐 Position formula: x = x₀ + v·t x = current position x₀ = initial position v = (constant) velocity t = time 💡 Examples: • Vehicle on cruise control on a straight highway • A train at steady speed • A body in space far from any forces |
Question 10
2.00 pts
📚 Concept summary:
Which of the following variables is a vector?
Explanation:
💡 Detailed explanation:
Summary - vectors and scalars! 📚
✓ Vectors in kinematics: All kinematic variables are vectors! • Position (x) - has direction • Displacement (Δx) - has direction • Velocity (v) - has direction • Acceleration (a) - has direction 📊 Summary table:
💡 In 1D motion: Direction is just sign (+/-). In 2D: full vectors with components. |
Question 11
2.00 pts
📐 Equation 1:
What is the velocity equation for motion at constant acceleration?
Explanation:
💡 Detailed explanation:
Velocity equation! 📐
Equation 1 - velocity: v = v₀ + at Where: • v = final velocity • v₀ = initial velocity • a = (constant!) acceleration • t = time 🔍 Derivation: From the definition of acceleration: a = (v - v₀)/t Multiply both sides by t: at = v - v₀ Rearrange: v = v₀ + at ✓ 💡 Example: A vehicle accelerates from 10 m/s at acceleration a=2 m/s² for 5 seconds v = 10 + 2×5 v = 10 + 10 v = 20 m/s ⚠️ Caution: This equation is only valid when a is constant! If a changes, you need integration. |
Question 12
2.00 pts
📐 Equation 2:
What is the position equation for motion at constant acceleration?
Explanation:
💡 Detailed explanation:
Position equation! 📐
Equation 2 - position: x = x₀ + v₀t + ½at² Where: • x = final position • x₀ = initial position • v₀ = initial velocity • a = acceleration • t = time 🔍 Understanding the equation: x₀ - the starting point v₀t - the distance that would be covered at the initial velocity ½at² - the additional distance due to acceleration 💡 Example: A body starts at x₀=0 at velocity v₀=5 m/s at acceleration a=2 m/s² After t=4s: x = 0 + 5×4 + ½×2×16 x = 0 + 20 + 16 x = 36m ⭐ Sometimes used: For displacement: Δx = v₀t + ½at² (when x₀ = 0) |
Question 13
2.00 pts
📐 Equation 3:
What is the equation linking v, v₀, a, and Δx without t?
Explanation:
💡 Detailed explanation:
Timeless equation! 📐
Equation 3 - without time: v² = v₀² + 2aΔx or: v² = v₀² + 2a(x-x₀) Where: Δx = x - x₀ (displacement) 🔍 Derivation: From equation 1: v = v₀ + at → t = (v-v₀)/a From equation 2: x = x₀ + v₀t + ½at² Substitute t: x - x₀ = v₀(v-v₀)/a + ½a[(v-v₀)/a]² After simplification: v² = v₀² + 2a(x-x₀) ✓ 💡 When to use it? When you don't know t and don't need it. Example: ball falls from height 5m, what is the velocity on impact? v² = 0 + 2×10×5 = 100 v = 10 m/s ✓ |
Question 14
2.00 pts
📋 Formula summary:
How many main formulas exist for motion at constant acceleration?
Explanation:
💡 Detailed explanation:
The formula table! 📋
📚 The 3 main formulas:
📐 Bonus formula: Average velocity (constant a): v̄ = (v + v₀)/2 💡 Solution strategy: 1. List what you know 2. List what's asked 3. Choose the formula with those variables 4. Solve algebraically |
Question 15
2.00 pts
🧮 Exercise:
A vehicle accelerates from v₀=10 m/s at acceleration a=2 m/s²
for t=5s
What is the final velocity?
Explanation:
💡 Detailed explanation:
Step-by-step solution! 🧮
| 📐 Solution: Given: • v₀ = 10 m/s • a = 2 m/s² • t = 5s Find: v = ? Choosing a formula: Known: v₀, a, t Looking for: v → Equation 1! v = v₀ + at Calculation: v = 10 + 2×5 v = 10 + 10 v = 20 m/s 💡 Sanity check: Reasonable? Yes! • Started at 10 m/s • Accelerates at 2 m/s² (each second +2 m/s) • After 5 seconds: +10 m/s • Total: 20 m/s ✓ |
Question 16
2.00 pts
🧮 Exercise:
A vehicle starts from rest (v₀=0) and accelerates at a=3 m/s²
for t=4s
What distance does it cover?
Explanation:
💡 Detailed explanation:
Distance traveled! 🧮
| 📐 Solution: Given: • v₀ = 0 (starts from rest!) • a = 3 m/s² • t = 4s • x₀ = 0 (assume) Find: Δx = ? Formula: x = x₀ + v₀t + ½at² Since v₀=0: x = x₀ + ½at² Calculation: x = 0 + ½×3×4² x = ½×3×16 x = ½×48 x = 24m Δx = x - x₀ = 24 - 0 = 24m 💡 Insight: Why isn't it just a×t? Because the velocity grows over time. Average velocity is half of the final velocity (since v₀=0): v̄ = (0+12)/2 = 6 m/s Δx = v̄ × t = 6 × 4 = 24m ✓ |
Question 17
2.00 pts
🧮 Exercise:
A vehicle accelerates from v₀=5 m/s to v=15 m/s
over Δx=50m
What is the acceleration?
Explanation:
💡 Detailed explanation:
Acceleration calculation! 🧮
| 📐 Solution: Given: • v₀ = 5 m/s • v = 15 m/s • Δx = 50m Find: a = ? Formula: Known: v₀, v, Δx Looking for: a No t → Equation 3! v² = v₀² + 2aΔx Calculation: 15² = 5² + 2×a×50 225 = 25 + 100a 200 = 100a a = 200/100 a = 2 m/s² 💡 Verification: Can find t: v = v₀ + at 15 = 5 + 2t t = 5s And check with equation 2: x = 0 + 5×5 + ½×2×25 x = 25 + 25 = 50m ✓ |
Question 18
2.00 pts
🛑 Deceleration:
A vehicle moves at v₀=20 m/s and brakes at a=-4 m/s²
How long until it stops?
Explanation:
💡 Detailed explanation:
Stopping time! 🛑
| 📐 Solution: Given: • v₀ = 20 m/s • a = -4 m/s² (negative = deceleration!) • v = 0 (stops!) Find: t = ? Formula: v = v₀ + at Calculation: 0 = 20 + (-4)×t 0 = 20 - 4t 4t = 20 t = 20/4 t = 5s 💡 Insight: The velocity decreases by 4 m/s every second: t=0: v=20 m/s t=1: v=16 m/s t=2: v=12 m/s t=3: v=8 m/s t=4: v=4 m/s t=5: v=0 m/s ✓ ⚠️ Note: Negative acceleration sign means motion is opposed. Important: the vehicle will continue to move (in reverse) if a stays negative beyond t=5s — physically, friction stops the vehicle once at rest. |
Question 19
2.00 pts
🛑 Braking distance:
A vehicle moves at v₀=30 m/s and brakes at a=-5 m/s²
What distance does it cover until it stops?
Explanation:
💡 Detailed explanation:
Braking distance! 🛑
| 📐 Solution: Given: • v₀ = 30 m/s • a = -5 m/s² • v = 0 (stops!) Find: Δx = ? Formula: No t? Equation 3! v² = v₀² + 2aΔx Calculation: 0² = 30² + 2×(-5)×Δx 0 = 900 - 10Δx 10Δx = 900 Δx = 900/10 Δx = 90m 💡 Verification: Can find t: 0 = 30 - 5t → t = 6s Average velocity: v̄ = (30+0)/2 = 15 m/s Distance: 15×6 = 90m ✓ ⚠️ Real-world note: Braking distance grows with v²! Double speed → 4× braking distance. That's why speed limits matter so much for road safety. |
Question 20
2.00 pts
🧮 Combined exercise:
A vehicle accelerates from rest at a=2 m/s² for 10s,
then moves at constant velocity for an additional 5s
What is the total distance covered?
Explanation:
💡 Detailed explanation:
Combined exercise! 🧮
| 📐 Full solution: Phase 1: Acceleration (t=0 to t=10s) • v₀ = 0 • a = 2 m/s² • t = 10s Final velocity: v = 0 + 2×10 = 20 m/s Distance: Δx₁ = ½×2×10² Δx₁ = 1×100 Δx₁ = 100m Phase 2: Constant velocity (t=10s to t=15s) • v = 20 m/s (constant!) • a = 0 • t = 5s Distance: Δx₂ = v×t Δx₂ = 20×5 Δx₂ = 100m Total: Δx = Δx₁ + Δx₂ Δx = 100 + 100 Δx = 200m 💡 Method: For multi-phase motion: handle each phase separately, then sum. |
Question 21
2.00 pts
📊 x-t graph:
What does the position-time graph look like for uniform motion (constant velocity)?
Explanation:
💡 Detailed explanation:
x-t graph for uniform motion! 📊
| 📈 Position-time graph: Uniform motion: x = x₀ + vt This is the equation of a straight line! • y-intercept: x₀ • Slope: v (the velocity!) 📊 Example: x = 10 + 5t • Starts at x₀=10m • Velocity v=5 m/s The line goes up steadily, with slope 5. 📐 Reading the slope: slope = Δx/Δt = velocity! • Steep line = fast • Shallow line = slow • Horizontal line = at rest (v=0) • Negative slope = moving in negative direction |
Question 22
2.00 pts
📊 x-t graph:
What does the position-time graph look like for motion at constant acceleration?
Explanation:
💡 Detailed explanation:
x-t graph with acceleration! 📊
| 📈 Graph with acceleration: Accelerated motion: x = x₀ + v₀t + ½at² Has t² = quadratic equation! → parabola 📊 Example: x = 0 + 0 + ½×2×t² = t² The graph curves upward. 📐 Properties of the parabola: Direction of curvature: • a > 0 → opens upward (concave up) • a < 0 → opens downward (concave down) Slope of the curve = velocity: • Tangent at any point = instantaneous velocity at that point • Steepening curve = velocity growing • Flattening curve = velocity decreasing |
Question 23
2.00 pts
📊 v-t graph:
What does the area under the curve in a velocity-time graph represent?
Explanation:
💡 Detailed explanation:
Area in the v-t graph! 📊
| 📐 An important principle: In the v-t graph: Area = displacement (Δx) Why? Δx = v × Δt That's exactly the area of a rectangle! 📊 Example - uniform motion: v = 10 m/s (constant) t = 5s Area = 10 × 5 = 50 So Δx = 50m ✓ 📐 In acceleration: If the graph is a tilted line (constant a): The area is a triangle or trapezoid Triangle (v₀=0): Area = ½×base×height = ½×t×v Trapezoid (v₀≠0): Area = ½×(v₀+v)×t 💡 Generalization: For any v(t) curve: Δx = ∫v dt (integral of v over t) = area under the curve. |
Question 24
2.00 pts
📊 v-t graph:
What does the slope in the velocity-time graph represent?
Explanation:
💡 Detailed explanation:
Slope in the v-t graph! 📊
| 📐 The principle: In the v-t graph: Slope = acceleration (a) Why? a = Δv/Δt That is the definition of slope! 📊 Example: v changes from 10 to 20 m/s in 5s slope = (20-10)/5 = 2 m/s² So a = 2 m/s² ✓ 📐 Reading the slope: By slope: • Positive slope = positive acceleration (speeding up) • Negative slope = negative acceleration (slowing down) • Horizontal line (slope=0) = constant velocity (a=0) • Steep slope = large acceleration • Shallow slope = small acceleration 💡 Summary table:
|
Question 25
2.00 pts
📊 a-t graph:
What does the acceleration-time graph look like for motion at constant acceleration?
Explanation:
💡 Detailed explanation:
a-t graph! 📊
| 📈 Acceleration graph: Constant acceleration: a = constant → horizontal line (straight) 📊 Example: a = 2 m/s² (constant) The graph is a flat horizontal line at the value a=2. 📐 Other types of acceleration:
💡 Note: In the a-t graph, the area under the line equals the change in velocity (Δv). |
Question 26
2.00 pts
📊 Graph analysis:
In a v-t graph, the line starts horizontal, then rises, and is again horizontal.
Describe the motion:
Explanation:
💡 Detailed explanation:
Analysis of a complex graph! 📊
| 📈 The graph: Three phases visible in the v-t graph. 📐 Analysis: Phase 1: Horizontal line v constant → a = 0 → Uniform motion (constant velocity) Phase 2: Rising tilted line v increasing → a > 0 → Acceleration (speeding up) Phase 3: Higher horizontal line v constant (and higher) → a = 0 → New uniform motion (faster) 💡 Real-world example: Vehicle on highway: 1. Cruising at 60 km/h 2. Driver accelerates to 100 km/h 3. Cruising at the new speed (100) ⭐ Tip: To read motion from a graph: 1. Identify the line shape in each phase 2. Translate to acceleration and velocity 3. Describe the motion in words |
Question 27
2.00 pts
📊 v-t graph:
In a v-t graph, the line is below the time axis (v negative).
What does this mean?
Explanation:
💡 Detailed explanation:
Negative velocity in the graph! 📊
| 📈 Negative velocity: v < 0: The body moves in the negative direction • x-axis: leftward • y-axis: downward 📊 Important distinction: Negative velocity ≠ Deceleration! • Negative v: direction of motion (sign matters) • Negative a: velocity decreasing These can be combined in 4 ways! 💡 Combinations:
|
Question 28
2.00 pts
📊 Graph exercise:
In an x-t graph, the body goes from x=20m to x=50m
between t=2s and t=8s
What is the velocity?
Explanation:
💡 Detailed explanation:
Velocity from x-t graph! 📊
| 📐 Solution: Velocity = slope of x-t! v = Δx/Δt Data from the graph: • At time t₁=2s: x₁=20m • At time t₂=8s: x₂=50m Calculation: Δx = x₂ - x₁ = 50 - 20 = 30m Δt = t₂ - t₁ = 8 - 2 = 6s v = 30/6 = 5 m/s 💡 Insight: Whenever the question is "find v from an x-t graph": → It's a slope question → Use Δx/Δt ⭐ Reverse: If the question is "find a from a v-t graph": → Slope question too → Use Δv/Δt |
Question 29
2.00 pts
📊 Graph exercise:
In a v-t graph, a body moves at constant velocity 15 m/s
for 4 seconds
What is the displacement?
Explanation:
💡 Detailed explanation:
Displacement from v-t graph! 📊
| 📐 Solution: Displacement = area under the v-t graph! Δx = v × Δt Calculation: v = 15 m/s Δt = 4s Δx = 15 × 4 = 60m Visualization: The graph is a horizontal line at v=15. The area beneath it from t=0 to t=4 is a rectangle: 15 × 4 = 60 💡 Note: If the question were about constant acceleration (line not horizontal): The area would be a triangle or trapezoid. ⭐ The principle: "Find Δx from v-t graph" → area under the curve! "Find Δv from a-t graph" → area under the curve! |
Question 30
2.00 pts
📚 Graphs summary:
What is the best way to identify acceleration from graphs?
Explanation:
💡 Detailed explanation:
Identifying acceleration from graphs! 📚
| 🔍 How to identify acceleration? ✓ From x-t graph: • Straight line → a = 0 (uniform motion) • Curve/parabola → a ≠ 0 (acceleration!) Parabola opening up ∪ → a > 0 Parabola opening down ∩ → a < 0 ✓ From v-t graph: • Horizontal line → a = 0 • Slope → a ≠ 0 Positive slope ↗ → a > 0 Negative slope ↘ → a < 0 The slope = the magnitude of the acceleration! ✓ From a-t graph: • Direct! a is the y-axis value • Just read it ⭐ Insight: v-t graph is usually the most informative (slope = a, area = Δx) |
Question 31
2.00 pts
🌍 Gravity:
What is the magnitude of gravitational acceleration near Earth's surface?
Explanation:
💡 Detailed explanation:
Gravitational acceleration! 🌍
Gravitational acceleration: g = 9.8 m/s² or approximately: g ≈ 10 m/s² Direction: downward (toward Earth's center) 🔍 Meaning: Any body in free fall: • Its velocity grows by 9.8 m/s • Every second! t=0: v=0 t=1: v=9.8 m/s t=2: v=19.6 m/s t=3: v=29.4 m/s ⋮ 💡 Important facts: ✓ g varies slightly by latitude (9.78 to 9.83) ✓ g decreases with height above sea level ✓ On the Moon: g ≈ 1.6 m/s² ✓ On Mars: g ≈ 3.7 m/s² ✓ In space (zero g): no apparent weight ⚠️ Note: g is independent of the body's mass! A feather and a hammer fall the same in vacuum. |
Question 32
2.00 pts
🪂 Free fall:
What is free fall?
Explanation:
💡 Detailed explanation:
Free fall! 🪂
Free fall: Motion under gravity only Properties: • Acceleration: a = g (constant!) • No air resistance • No other forces 🔍 Formulas: Exactly like motion at constant acceleration, just replace a with g: 1️⃣ v = v₀ + gt 2️⃣ y = y₀ + v₀t + ½gt² 3️⃣ v² = v₀² + 2gΔy 💡 Examples of free fall: • Apple from a tree • Stone thrown up • Skydiver before opening parachute (approximate) • Astronauts in orbit (Earth in free fall too) ⚠️ Sign convention: If we choose up = positive: g = -9.8 m/s² (downward = negative) If we choose down = positive: g = +9.8 m/s² |
Question 33
2.00 pts
🧮 Fall exercise:
A stone falls from a height (v₀=0)
after t=3s, what is its velocity? (g=10 m/s²)
Explanation:
💡 Detailed explanation:
Fall from rest! 🧮
| 📐 Solution: Given: • v₀ = 0 (falls from rest!) • g = 10 m/s² (downward) • t = 3s Find: v = ? Formula: v = v₀ + gt Calculation: v = 0 + 10×3 v = 30 m/s Direction: downward ↓ 💡 Insight: Each second the velocity grows by 10 m/s: t=0: v=0 t=1: v=10 m/s t=2: v=20 m/s t=3: v=30 m/s ✓ ⚠️ With signs: If up is positive: v = -30 m/s (down) |
Question 34
2.00 pts
🧮 Exercise:
A stone falls from rest for t=2s
What height has it covered? (g=10 m/s²)
Explanation:
💡 Detailed explanation:
Fall height! 🧮
| 📐 Solution: Given: • v₀ = 0 • g = 10 m/s² • t = 2s • y₀ = 0 (assume) Formula: y = y₀ + v₀t + ½gt² Since v₀=0: y = ½gt² Calculation: y = ½×10×2² y = 5×4 y = 20m 💡 Insight: The body fell 20 meters down If it started from height 100m: → now at height 80m ⚠️ Useful formula: For fall from rest: After t=1s: 5m fallen After t=2s: 20m fallen After t=3s: 45m fallen After t=4s: 80m fallen Pattern: y = 5t² (with g=10) |
Question 35
2.00 pts
🧮 Exercise:
A stone falls from height h=80m
How long does it take to reach the ground? (g=10 m/s²)
Explanation:
💡 Detailed explanation:
Fall time! 🧮
| 📐 Solution: Given: • h = 80m (height) • v₀ = 0 • g = 10 m/s² Formula: h = ½gt² Solution: 80 = ½×10×t² 80 = 5t² t² = 80/5 t² = 16 t = √16 t = 4s 💡 General formula: t = √(2h/g) Fall time from height h Verification: t = √(2×80/10) t = √(160/10) t = √16 t = 4s ✓ ⚠️ Note: The time doesn't depend on the mass! A heavy stone and a feather (in vacuum) hit the ground at the same time. |
Question 36
2.00 pts
🧮 Exercise:
A stone falls from height h=45m
What is its velocity on impact with the ground? (g=10 m/s²)
Explanation:
💡 Detailed explanation:
Impact velocity! 🧮
| 📐 Solution: Given: • h = 45m • v₀ = 0 • g = 10 m/s² Formula (without time!): v² = v₀² + 2gh Calculation: v² = 0² + 2×10×45 v² = 900 v = √900 v = 30 m/s 💡 General formula: v = √(2gh) Impact velocity from height h Interesting ratio: Doubling the height → multiplies impact velocity by √2 ≈ 1.41 Quadrupling the height → multiplies impact velocity by 2 |
Question 37
2.00 pts
🎯 Throw upward:
A ball is thrown upward at velocity v₀=20 m/s
What maximum height does it reach? (g=10 m/s²)
Explanation:
💡 Detailed explanation:
Maximum height! 🎯
| 📐 Solution: Given: • v₀ = +20 m/s (upward!) • g = -10 m/s² (downward!) • At the highest point: v = 0 Why v=0 at the highest point? The ball: • Rises and slows (g opposes) • Reaches the highest point • Stops momentarily (v=0) • Starts to fall Formula: v² = v₀² + 2gΔy Calculation: 0² = 20² + 2×(-10)×h 0 = 400 - 20h 20h = 400 h = 400/20 h = 20m 💡 Direct formula: h = v₀²/(2g) Max height for an upward throw |
Question 38
2.00 pts
🧮 Exercise:
A ball is thrown upward at v₀=30 m/s
How long does it take to reach the highest point? (g=10 m/s²)
Explanation:
💡 Detailed explanation:
Time to peak! 🧮
| 📐 Solution: Given: • v₀ = +30 m/s • g = -10 m/s² • v = 0 (at the highest point) Formula: v = v₀ + gt Calculation: 0 = 30 + (-10)×t 0 = 30 - 10t 10t = 30 t = 3s 💡 Direct formula: t_up = v₀/g Time to reach the highest point Total flight time: By symmetry: time up = time down T_total = 2 × t_up = 2 × 3 = 6s Then the ball returns to the original height at the same speed v₀ = 30 m/s (downward) |
Question 39
2.00 pts
🧮 Complex exercise:
From a roof at height 20m, a ball is thrown upward at 10 m/s
What is the velocity on impact with the ground? (g=10 m/s²)
Explanation:
💡 Detailed explanation:
Throw from a height! 🧮
| 📐 Solution: Given: • y₀ = +20m (roof height) • v₀ = +10 m/s (upward) • y = 0 (ground) • g = -10 m/s² Formula: v² = v₀² + 2g(y-y₀) Calculation: Δy = 0 - 20 = -20m (fell 20m below the start) v² = 10² + 2×(-10)×(-20) v² = 100 + 400 v² = 500 v = √500 v ≈ 22.4 m/s 💡 Step understanding: Phase 1: ball rises from roof up to peak (slowing from 10 to 0 m/s) Phase 2: ball falls all the way to the ground (speeding up from 0 to ~22.4 m/s) ⭐ Key insight: The ball's velocity on impact only depends on: • Initial speed (magnitude only!) • Total height drop Direction of initial throw doesn't affect the final speed (energy conservation). |
Question 40
2.00 pts
📚 Free fall summary:
What is the central condition for free fall?
Explanation:
💡 Detailed explanation:
Free fall summary! 📚
📋 Central formulas:
⭐ Properties of free fall: ✓ Constant acceleration: a = g = 9.8 m/s² ✓ Independent of mass ✓ Independent of shape (in vacuum) ✓ Always toward Earth's center 💡 In real life: For most everyday objects (stones, balls, etc.) air resistance is small enough that the free-fall model gives accurate predictions. ⚠️ When air drag matters: • Light objects (feathers, paper) • Large surface area (parachute) • Very high velocities (terminal velocity) | ||||||||
Question 41
2.00 pts
🚗 Combined exercise:
A vehicle accelerates from 0 to 20 m/s in 4s,
moves at constant velocity for 6s,
then brakes to a stop in 5s
What is the total distance covered?
Explanation:
💡 Detailed explanation:
Combined motion exercise! 🚗
| 📐 Step-by-step solution: Phase 1: Acceleration (0→4s) • v₀ = 0 • v = 20 m/s • t = 4s • a = (20-0)/4 = 5 m/s² Distance: Δx₁ = ½×5×4² Δx₁ = ½×5×16 Δx₁ = 40m Or: average velocity = 10 m/s Δx₁ = 10×4 = 40m ✓ Phase 2: Constant velocity (4→10s) • v = 20 m/s (constant!) • t = 6s • a = 0 Distance: Δx₂ = v×t Δx₂ = 20×6 Δx₂ = 120m Phase 3: Braking (10→15s) • v₀ = 20 m/s • v = 0 • t = 5s • a = -4 m/s² Distance (avg velocity = 10 m/s): Δx₃ = 10×5 = 50m Total: Δx = 40 + 120 + 50 = 210m |
Question 42
2.00 pts
🚗🏃 Two bodies:
A vehicle moves at constant velocity 15 m/s
A runner accelerates from rest at 3 m/s²
When does the runner catch the vehicle? (both from the same point)
Explanation:
💡 Detailed explanation:
Catch-up problem! 🚗🏃
| 📐 Solution: Vehicle (constant velocity): x₁ = 15t Runner (accel from 0): x₂ = ½×3×t² = 1.5t² Catch-up condition: x₁ = x₂ 15t = 1.5t² Solution: 1.5t² = 15t 1.5t² - 15t = 0 1.5t(t - 10) = 0 Solutions: • t = 0 (start) • t = 10s (catches up!) 💡 Verification: After 10 seconds: Vehicle: x₁ = 15×10 = 150m Runner: x₂ = 1.5×100 = 150m ✓ Runner's speed: v = 3×10 = 30 m/s (2× the vehicle!) ⭐ Insight: The runner is initially slower but catches up because they keep accelerating. By the time of meeting, they're going twice as fast as the vehicle. |
Question 43
2.00 pts
🚗 Road safety:
A vehicle moves at velocity v and brakes.
If the velocity is doubled,
the braking distance will be:
Explanation:
💡 Detailed explanation:
Braking distance and velocity! 🚗
| 📐 Explanation: Braking distance: From v² = v₀² + 2aΔx When stopping (v=0): 0 = v₀² - 2aΔx Δx = v₀²/(2a) Braking distance is proportional to the square of the velocity! 💡 Numerical example:
⚠️ Importance for road safety: 50 km/h → 25m to stop 100 km/h → 100m to stop (quadruple, not double!) That's why speed limits matter so much. |
Question 44
2.00 pts
🪨 Two stones:
Stone A falls from rest.
One second later, stone B falls from the same place.
The distance between them:
Explanation:
💡 Detailed explanation:
Two falling stones! 🪨
| 📐 Analysis: Stone A at time t: y_A = ½gt² v_A = gt Stone B at time t: (started one second later!) y_B = ½g(t-1)² v_B = g(t-1) Distance between them: Δy = y_A - y_B = ½gt² - ½g(t-1)² = ½g[t² - (t-1)²] = ½g[t² - t² + 2t - 1] = ½g(2t - 1) = gt - ½g Conclusion: The distance depends on t - grows over time! Example (g=10): t=2s: Δy = 10×2 - 5 = 15m t=3s: Δy = 10×3 - 5 = 25m t=4s: Δy = 10×4 - 5 = 35m 💡 Why? A is always 1 second ahead, but during that second its velocity grew. So A is always faster than B → the gap keeps widening. |
Question 45
2.00 pts
⚖️ Units:
A vehicle accelerates from 0 to 108 km/h in 5 seconds.
What is the acceleration in m/s²?
Explanation:
💡 Detailed explanation:
Unit conversion! ⚖️
| 📐 Solution: Step 1: Convert to m/s 108 km/h = ? 108 ÷ 3.6 = 30 m/s Step 2: Acceleration calculation a = Δv/Δt a = (30-0)/5 a = 30/5 a = 6 m/s² 💡 Remember:
⚠️ Critical: Always convert to SI units before computing! Mixing km/h with seconds gives wrong numbers. |
Question 46
2.00 pts
📊 Graph analysis:
In a v-t graph, the line rises, then is horizontal, then drops below the time axis.
Describe the motion:
Explanation:
💡 Detailed explanation:
Complex graph analysis! 📊
| 📈 Phase-by-phase analysis: Phase 1: Rising line (v: 0→positive) v > 0, slope > 0 → Moving right, speeding up (positive acceleration) Phase 2: Horizontal line (v positive) v > 0, slope = 0 → Moving right, constant velocity (no acceleration) Phase 3: Line drops to v = 0 v decreasing to 0, slope < 0 → Moving right, slowing down (negative acceleration, opposes motion) Phase 4: Line below the axis (v < 0) v < 0, slope < 0 → Moving left, speeding up in that direction! (negative acceleration in same direction as motion) ⭐ Key insight: The body changed direction at the moment v=0! (when the line crossed the time axis) |
Question 47
2.00 pts
⚠️ Common error:
Which of the following statements is wrong?
Explanation:
💡 Detailed explanation:
Common kinematics errors! ⚠️
| ❌ The error: Wrong claim: "If a is negative, the body slows down" Not always true! 🔍 Explanation: Negative acceleration means: • The acceleration points to the left That doesn't necessarily mean the body slows down! 💡 Examples:
⭐ The correct rule: The body slows down only when v and a have opposite signs. If they have the same sign → speeds up. |
Question 48
2.00 pts
💡 Solution tip:
What is the first step in solving a kinematics problem?
Explanation:
💡 Detailed explanation:
Problem-solving method! 💡
📋 5-step method: 1️⃣ List the data • What is given? • What is asked? • Mark known variables 2️⃣ Diagram (if needed) • Draw the situation • Mark directions • Set a coordinate system 3️⃣ Choose a formula • Which formula fits? • What variables does it have? 4️⃣ Calculate • Plug in values • Solve • Include units! 5️⃣ Sanity check • Is the answer reasonable? • Are the units right? • Is the sign correct? 💡 Why this works: Most errors come from skipping steps 1-2. Listing data and drawing a picture take 30 seconds and prevent most calculation mistakes. |
Question 49
2.00 pts
📚 Formula summary:
How many main formulas do you need to know?
Explanation:
💡 Detailed explanation:
Summary of all formulas! 📚
📋 The 3 basic formulas:
📐 Bonus formulas: Free fall: Replace a with g = 9.8 m/s² Average velocity (constant a): v̄ = (v₀ + v)/2 Definition: a = Δv/Δt, v = Δx/Δt 💡 The big idea: You don't need to memorize many formulas — just understand 3 well, and the rest follow! |
Question 50
2.00 pts
🎓 Summary of Exam 161:
What is the central topic of one-dimensional kinematics?
Explanation:
💡 Detailed explanation:
Summary of Exam 161 - 1D Kinematics! 🎓
🌟 Comprehensive summary: 1D Kinematics Description of straight-line motion (1D) 📚 What we learned: Part A: Foundational concepts ✓ Position, displacement, distance ✓ Velocity (average and instantaneous) ✓ Acceleration ✓ Vectors and scalars Part B: Formulas ✓ v = v₀ + at ✓ x = x₀ + v₀t + ½at² ✓ v² = v₀² + 2aΔx ✓ Average velocity Part C: Graphs ✓ x-t, v-t, a-t graphs ✓ Slope = derivative ✓ Area = integral Part D: Free fall ✓ g = 9.8 m/s² ✓ Same formulas with g ✓ Up: ball stops at peak (v=0) 💡 Central takeaway: 3 formulas describe all uniform-acceleration motion! Choose the formula based on which variables you have and which you need. Free fall is just the special case with a = g. |