2
Chapter 2· Kinematics
Numericals Solved Step by Step
Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.
Numerical 2.1
Represent (a) a velocity of 400 m s⁻¹ at 60° to the x-axis and (b) a force of 50 N at 120° to the x-axis.
Given Data
- Scale for (a): 1 cm = 100 m s⁻¹
- Scale for (b): 1 cm = 10 N
To Find
- Length of each arrow, and the drawing
Formula
length = magnitude ÷ scale
Calculation
- (a) Length = 400 ÷ 100 = 4 cm. Draw a 4 cm line at 60° from the positive x-axis and add an arrowhead.
- (b) Length = 50 ÷ 10 = 5 cm. Draw a 5 cm line at 120° from the positive x-axis (it points up and to the left) and add an arrowhead.
Result(a) a 4 cm arrow at 60° (b) a 5 cm arrow at 120°
Note: Any sensible scale is fine, as long as you state it and the length matches it.
Numerical 2.2
A car is moving at an average speed of 72 km h⁻¹. How much time will it take to cover a distance of 360 km?
Formula
v = S ÷ t, so t = S ÷ v
Calculation
- From v = S ÷ t, the time t = S ÷ v.
- t = 360 km ÷ 72 km h⁻¹ = 5 h.
- Here both values are already in km and hours, so no conversion is needed.
Numerical 2.3
A truck starts from rest and reaches a velocity of 90 km h⁻¹ in 50 s. Find its average acceleration.
Given Data
- vᵢ = 0
- vf = 90 km h⁻¹ = 90 ÷ 3.6 = 25 m s⁻¹
- t = 50 s
Calculation
- a = (vf − vᵢ) ÷ t.
- a = (25 − 0) ÷ 50.
Numerical 2.4
A car is moving at 5 m s⁻¹ and speeds up with an acceleration of 1.5 m s⁻². What is its velocity after 5 s?
Given Data
- vᵢ = 5 m s⁻¹
- a = 1.5 m s⁻²
- t = 5 s
Calculation
- vf = vᵢ + at.
- vf = 5 + 1.5 × 5 = 5 + 7.5.
Numerical 2.5
A motorcycle moving at 18 km h⁻¹ accelerates at 2 m s⁻². How far will it travel in 10 s?
Given Data
- vᵢ = 18 km h⁻¹ = 18 ÷ 3.6 = 5 m s⁻¹
- a = 2 m s⁻²
- t = 10 s
Calculation
- S = vᵢt + ½at².
- S = 5 × 10 + ½ × 2 × 10².
- S = 50 + 100.
Numerical 2.6
A wagon moving at 54 km h⁻¹ is brought to rest within 25 m. Find its acceleration.
Given Data
- vᵢ = 54 km h⁻¹ = 54 ÷ 3.6 = 15 m s⁻¹
- vf = 0 (comes to rest)
- S = 25 m
Calculation
- Time is not given, so use 2aS = vf² − vᵢ².
- 2 × a × 25 = 0² − 15² = −225.
- a = −225 ÷ 50.
Resulta = −4.5 m s⁻² (a deceleration of 4.5 m s⁻²)
Note: The minus sign shows the acceleration is opposite to the motion: the wagon is slowing down.
Numerical 2.7
A stone is dropped from a height of 45 m. How long will it take to reach the ground, and with what velocity will it hit the ground?
Given Data
- vᵢ = 0
- h = 45 m
- g = 10 m s⁻²
To Find
- Time t to reach the ground
- Velocity vf on hitting the ground
Formulas
h = vᵢt + ½gt²
vf = vᵢ + gt
Calculation
- Time: h = vᵢt + ½gt² gives 45 = ½ × 10 × t², so t² = 9 and t = 3 s.
- Velocity: vf = vᵢ + gt = 0 + 10 × 3 = 30 m s⁻¹.
- Check: 2gh = 2 × 10 × 45 = 900 = 30², so vf = 30 m s⁻¹.
Resultt = 3 s, vf = 30 m s⁻¹
Numerical 2.8
A car travels 10 km at an average speed of 20 m s⁻¹ and then 0.8 km at an average speed of 4 m s⁻¹. Find its average speed for the whole journey.
Given Data
- S₁ = 10 km = 10 000 m at v₁ = 20 m s⁻¹
- S₂ = 0.8 km = 800 m at v₂ = 4 m s⁻¹
To Find
- Average speed for the whole journey
Formulas
t = S ÷ v
average speed = total distance ÷ total time
Calculation
- Time for the first part: t₁ = 10 000 ÷ 20 = 500 s.
- Time for the second part: t₂ = 800 ÷ 4 = 200 s.
- Average speed = total distance ÷ total time = (10 000 + 800) ÷ (500 + 200) = 10 800 ÷ 700.
ResultAverage speed ≈ 15.4 m s⁻¹
Note: The average is not (20 + 4) ÷ 2 = 12 m s⁻¹: the car spends different times at each speed, so always divide total distance by total time.
Numerical 2.9
A ball is dropped from the top of a building and takes 5 s to reach the ground. Find the height of the building and the velocity with which the ball hits the ground.
Given Data
- vᵢ = 0
- t = 5 s
- g = 10 m s⁻²
To Find
- Height h of the building
- Velocity vf on hitting the ground
Formulas
h = vᵢt + ½gt²
vf = vᵢ + gt
Calculation
- Height: h = vᵢt + ½gt² = 0 + ½ × 10 × 5² = 125 m.
- Velocity: vf = vᵢ + gt = 0 + 10 × 5 = 50 m s⁻¹.
Resulth = 125 m, vf = 50 m s⁻¹
Numerical 2.10
A ball is thrown straight up and takes 3 s to reach its highest point. Find its initial velocity and the maximum height it reaches above the ground, if it was released 1 m above the ground.
Given Data
- t = 3 s to the top
- vf = 0 at the top
- g = −10 m s⁻² (going up)
- released 1 m above the ground
To Find
- Initial velocity vᵢ
- Maximum height above the ground
Formulas
vf = vᵢ + gt
2gh = vf² − vᵢ²
Calculation
- Initial velocity: vf = vᵢ + gt gives 0 = vᵢ − 10 × 3, so vᵢ = 30 m s⁻¹.
- Height above the release point: 2gh = vf² − vᵢ² gives 2(−10)h = 0 − 30², so h = 900 ÷ 20 = 45 m.
- Height above the ground = 45 m + 1 m = 46 m.
Resultvᵢ = 30 m s⁻¹, maximum height = 46 m above the ground
