Rectilinear Acceleration – SS1 Physics Lesson Note (First Term)

Rectilinear acceleration is the rate of change of velocity with time for a body moving in a straight line.

• Unit: metre per second squared (m/s²).
• Uniform acceleration: velocity changes by equal amounts in equal intervals of time.
• Equations of motion: three equations link u, v, a, t and S.

Behavioural objectives

By the end of the lesson, students should be able to:

1. define acceleration and state its unit;
2. explain what a negative acceleration (deceleration or retardation) means;
3. describe uniform acceleration and acceleration due to gravity;
4. derive the three equations of uniformly accelerated motion;
5. use the equations to solve problems, following the correct steps and units.

What is acceleration?

Acceleration is the rate of change of velocity with time.

• Accelerated body: any body whose velocity changes with time is said to be accelerated.
• Unit: m/s², because velocity is in m/s and it is divided by time in seconds.
• Type of quantity: acceleration is a vector quantity, so it has size and direction.

Suppose a car moves with velocity u and then increases its velocity to v in a time interval t. Its acceleration is:

a = (v – u)/t

• a: acceleration in m/s².
• v: final velocity in m/s.
• u: initial velocity in m/s.
• t: time taken for the change, in seconds.

What is negative acceleration (deceleration)?

Negative acceleration is a decrease in velocity with time, also called deceleration or retardation.

• Cause of the negative sign: if a car slows down from v to u, then (v – u) is negative, so a is negative.
• Meaning: the body is slowing down.
• Example: a bus braking as it approaches a bus stop.

What are the possible cases of acceleration?

There are three cases: positive acceleration, negative acceleration and zero acceleration.

What is uniform acceleration?

Uniform acceleration is motion in which the velocity of a body increases by equal amounts in equal intervals of time, however small the interval.

How do you read the velocity-time graph?

• Straight sloping line: the acceleration is uniform.
• Slope: the slope of the line equals the acceleration, a = (v – u)/t.
• Area under the line: the area between the line and the time axis equals the distance covered, S.
• Intercept on the velocity axis: this is the initial velocity, u.

What is acceleration due to gravity?

Acceleration due to gravity (g) is the constant acceleration of a body falling freely near the earth’s surface, about 9.8 m/s².

• Falling body: its velocity increases steadily from zero, so it accelerates.
• Body thrown upwards: its speed decreases gradually until it stops at the highest point (deceleration).
• Cause: the earth attracts bodies close to it or on its surface.
• Size: the velocity of a freely falling body increases by about 9.8 m/s every second.
• Variation: the value changes slightly from one place on the earth to another.
• Use in problems: use the value given in the question, often 10 m/s² or 9.8 m/s².

What sign should g have in calculations?

• Downward positive: a falling body has a = +g and a rising body has a = -g.
• Upward positive: a rising body has a = +g in size but the sign is reversed, so choose one direction and keep to it.

What are the equations of uniformly accelerated motion?

There are three equations of uniformly accelerated motion, and they apply only to straight-line motion with constant acceleration.

How is the 1st equation derived?

1. Start from the definition: a = (v – u)/t.
2. Multiply both sides by t: at = v – u.
3. Add u to both sides.
4. Result: v = u + at

How is the 2nd equation derived?

1. For a body whose velocity changes from u to v, the average velocity is (u + v)/2.
2. Distance S = average velocity × time, so S = ((u + v)/2)t.
3. Put v = u + at into this: S = ((u + u + at)/2)t.
4. Simplify: S = ((2u + at)/2)t.
5. Expand: S = 2ut/2 + at²/2.
6. Result: S = ut + ½at²

How is the 3rd equation derived?

1. Square both sides of the 1st equation: v² = (u + at)².
2. Expand: v² = u² + 2uat + a²t².
3. Factorise the last two terms: v² = u² + 2a(ut + ½at²).
4. Replace (ut + ½at²) with S, from the 2nd equation.
5. Result: v² = u² + 2aS

How do you solve problems with the equations?

Memorise the equations and be able to derive them, then follow these steps.

1. Identify all the quantities in the problem: u, v, a, t and S.
2. Decide which quantities are given and which one is to be found.
3. Choose the equation that connects the given quantities to the unknown.
4. Change all quantities to S.I. units unless told otherwise.
5. Solve the equation for the unknown quantity.
6. Write the answer with the correct unit.

What unit conversions are common?

• km/h to m/s: multiply by 1000 and divide by 3600, so 72 km/h = 20 m/s.
• Minutes to seconds: multiply by 60.
• Warning: mixed units give a wrong answer.

Worked example 1: starting from rest

A car starts from rest and accelerates uniformly at 2 m/s² for 10 s. Find its final velocity and the distance covered.

• Given: u = 0, a = 2 m/s², t = 10 s.
• Final velocity: v = u + at = 0 + 2 × 10 = 20 m/s.
• Distance: S = ut + ½at² = 0 + ½ × 2 × 10² = 100 m.

Worked example 2: deceleration

A car moving at 30 m/s brakes uniformly and stops in 6 s. Find the acceleration and the stopping distance.

• Given: u = 30 m/s, v = 0, t = 6 s.
• Acceleration: a = (v – u)/t = (0 – 30)/6 = -5 m/s².
• Distance: S = ut + ½at² = 30 × 6 + ½ × (-5) × 6² = 180 – 90 = 90 m.

Worked example 3: free fall

A stone is dropped from rest and falls for 3 s. Take g = 10 m/s². Find its velocity and the distance fallen.

• Given: u = 0, a = g = 10 m/s², t = 3 s.
• Velocity: v = u + at = 0 + 10 × 3 = 30 m/s.
• Distance: S = ½at² = ½ × 10 × 3² = 45 m.

Worked example 4: time not given

A body starts from rest with acceleration 4 m/s² and covers 8 m. Find its final velocity.

• Given: u = 0, a = 4 m/s², S = 8 m.
• Equation: v² = u² + 2aS = 0 + 2 × 4 × 8 = 64.
• Answer: v = 8 m/s.

Evaluation

1. Define acceleration and state its unit.
2. What does a negative value of acceleration indicate?
3. When is a body said to move with uniform acceleration?
4. Starting from a = (v – u)/t, derive the equation v = u + at.
5. List the steps for solving problems with the equations of motion.

Assignment

1. Derive S = ut + ½at² from the first equation of motion. Then use it to find the distance covered by a car that starts from rest and accelerates uniformly at 3 m/s² for 8 s.