📐 Quick Reference:
v = u + at
s = ut + ½at²
v² = u² + 2as
where u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement
Problem 1.1Beginner
A car travels at a constant speed of 20 m/s. How far does it travel in 30 seconds?
Step-by-Step Solution:
Given: speed = 20 m/s, time = 30 s
Formula: distance = speed × time
Calculation: distance = 20 × 30 = 600 meters
✅ Answer: 600 meters
💡 Tip: When speed is constant, distance = speed × time is all you need!
Problem 1.2Beginner
A runner completes a 100-meter race in 12.5 seconds. What is their average speed?
Step-by-Step Solution:
Given: distance = 100 m, time = 12.5 s
Formula: speed = distance / time
Calculation: speed = 100 / 12.5 = 8 m/s
✅ Answer: 8 m/s
Problem 1.3Beginner
A ball is thrown upward with an initial velocity of 15 m/s. How long does it take to reach its highest point? (g = 10 m/s²)
Step-by-Step Solution:
Given: u = 15 m/s, v = 0 m/s (at highest point), a = -10 m/s² (deceleration)
Formula: v = u + at
Calculation: 0 = 15 + (-10)t
10t = 15
t = 15/10 = 1.5 seconds
✅ Answer: 1.5 seconds
🔍 Observation Connection: This explains why the ball seems to pause at the top!
Problem 2.1Intermediate
A train starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. It then continues at constant speed for 20 seconds. Finally, it decelerates uniformly at 4 m/s² until it stops. Find the total distance traveled.
Step-by-Step Solution:
Phase 1 (Acceleration):
u = 0, a = 2 m/s², t = 10 s
v = u + at = 0 + 2×10 = 20 m/s
s₁ = ut + ½at² = 0 + ½×2×100 = 100 m
Phase 2 (Constant speed):
v = 20 m/s, t = 20 s
s₂ = v×t = 20×20 = 400 m
Phase 3 (Deceleration):
u = 20 m/s, v = 0, a = -4 m/s²
Using v² = u² + 2as: 0 = 400 + 2(-4)s₃
8s₃ = 400
s₃ = 50 m
Total distance: s₁ + s₂ + s₃ = 100 + 400 + 50 = 550 m
✅ Answer: 550 meters
Problem 2.2Intermediate
A stone is dropped from a cliff. It takes 4 seconds to hit the ground. Find the height of the cliff and the velocity just before impact. (g = 10 m/s²)
Step-by-Step Solution:
Given: u = 0, t = 4 s, a = g = 10 m/s²
Height (s = ut + ½at²): s = 0 + ½×10×16 = 80 m
Final velocity (v = u + at): v = 0 + 10×4 = 40 m/s
✅ Answer: Height = 80 m, Velocity = 40 m/s downward
Problem 3.1Advanced
Two cars start from rest from the same point. Car A accelerates at 2 m/s² for 10 seconds and then continues at constant speed. Car B accelerates at 3 m/s² for 8 seconds and then continues at constant speed. Which car is ahead after 20 seconds? By how much?
Step-by-Step Solution:
Car A:
Phase 1 (0-10s): u=0, a=2, t=10
v_A = 0 + 2×10 = 20 m/s
s_A1 = 0×10 + ½×2×100 = 100 m
Phase 2 (10-20s): v=20 m/s, t=10
s_A2 = 20×10 = 200 m
Total s_A = 100 + 200 = 300 m
Car B:
Phase 1 (0-8s): u=0, a=3, t=8
v_B = 0 + 3×8 = 24 m/s
s_B1 = 0×8 + ½×3×64 = 96 m
Phase 2 (8-20s): v=24 m/s, t=12
s_B2 = 24×12 = 288 m
Total s_B = 96 + 288 = 384 m
Difference: 384 - 300 = 84 m
✅ Answer: Car B is ahead by 84 meters
Problem 3.2Advanced
A ball is thrown vertically upward with velocity 20 m/s from the top of a 50 m tall building. Find: (a) Maximum height reached above ground, (b) Time taken to reach ground, (c) Velocity with which it hits the ground. (g = 10 m/s²)
Step-by-Step Solution:
Part (a): Maximum height above building
At highest point, v = 0
v² = u² - 2gh (taking upward as positive, g negative)
0 = 400 - 2×10×h
20h = 400
h = 20 m above building
Height above ground = 50 + 20 = 70 m
Part (b): Time to reach ground
Total displacement = -50 m (from top to ground, downward)
Using s = ut - ½gt² (g negative for upward positive)
-50 = 20t - ½×10×t²
-50 = 20t - 5t²
5t² - 20t - 50 = 0
t² - 4t - 10 = 0
t = [4 ± √(16 + 40)]/2 = [4 ± √56]/2
t = [4 + 7.48]/2 = 5.74 s (taking positive)
Part (c): Velocity at ground
v = u - gt = 20 - 10×5.74
v = 20 - 57.4 = -37.4 m/s
Negative means downward, so 37.4 m/s downward
✅ Answer: (a) 70 m, (b) 5.74 s, (c) 37.4 m/s downward
🧠 Critical Thinking: Notice how the final velocity is greater than the initial velocity? Why? (Hint: It falls from higher than it started!)
🧠 Critical Thinking Challenge:
In Problem 3.2, if you double the initial velocity to 40 m/s, does the maximum height double? Why or why not? (Hint: Look at the formula v² = u² - 2gh)
Try to reason it out mathematically before calculating!