1.3.1Dynamics
Acceleration
5 min read 6 parts Lesson 8 of 22
Section 1 of 6What Acceleration Means
Whenever the velocity of an object changes, the object is accelerating.
Definition
- Acceleration is the change in velocity per second, so a large acceleration means the velocity is changing rapidly.
Vector
- Acceleration is a vector with units m/s², because velocity is a vector and its change has a direction.
Deceleration
- When an object slows down its final velocity is smaller than its initial velocity, giving a negative acceleration.
Not just speeding up
- Changing direction at constant speed is also acceleration, because the velocity vector changes.
Section 2 of 6The Equation
The Equation
Relationship
- The acceleration is found from a = (v − u) ÷ t, where v is final velocity, u is initial velocity and t is time.
Symbols
- a is in m/s², v and u are in m/s, and t is the time taken for the change in seconds.
Change in velocity
- The top of the fraction (v − u) is the change in velocity, found by subtracting initial from final.
Rearranged
- This gives v = u + a t for the final velocity, and t = (v − u) ÷ a for the time needed for a change.
| Quantity | Symbol | Unit |
|---|---|---|
| acceleration | a | m/s² |
| final velocity | v | m/s |
| initial velocity | u | m/s |
| time | t | s |
Section 3 of 6Using It Step by Step
1
Identify the initial velocity u, which is zero if the object starts from rest.
2
Identify the final velocity v, which is zero if the object comes to a stop.
3
Convert velocities to m/s and time to seconds before substituting.
4
Subtract u from v to find the change in velocity, keeping the sign.
5
Divide by the time and give the unit m/s²; a negative answer means deceleration.
Section 4 of 6Everyday Examples
Everyday Examples
Sprinter
- A sprinter leaving the blocks changes velocity from 0 to about 10 m/s in roughly 2 s, an acceleration of around 5 m/s².
Braking car
- A car braking hard can slow by more than 6 m/s every second, so its acceleration is large and negative.
Lift
- A lift accelerates gently for comfort, then travels at constant velocity, then decelerates to stop at a floor.
Circular motion
- A satellite in a circular orbit has constant speed but still accelerates, because its direction continually changes.
Section 5 of 6Worked Example
Question
- A car increases its velocity from 5.0 m/s to 20 m/s in 6.0 s.
- Then it brakes from 30 m/s to rest in 10 s.
- Calculate both accelerations.
- First stagea = (v − u) ÷ t = (20 − 5.0) ÷ 6.0 = 15 ÷ 6.0 = 2.5 m/s².What the symbols mean: a = acceleration (m s−2) · v = final velocity · u = initial velocity (m s−1) · Δv = change in velocity · t = time taken (s)
- Second stagea = (0 − 30) ÷ 10 = −3.0 m/s².
- The negative sign shows deceleration; the car loses 3.0 m/s of velocity every second.
Answer
- Acceleration 2.5 m/s²;
- deceleration −3.0 m/s² (3.0 m/s² opposite to the motion).
Section 6 of 6Exam Checklist & Tips
1.3.1 – Key Points for the Exam
- Acceleration The change in velocity per unit time, in m/s².
- Initial velocity The velocity at the start of the time interval, u.
- Final velocity The velocity at the end of the time interval, v.
- Deceleration A negative acceleration where an object slows down.
Common Mistakes
- Subtracting the wrong way round, using (u − v) instead of (v − u).
- Giving the unit as m/s rather than m/s².
- Forgetting that an object starting from rest has u = 0.
- Thinking that a moving object must have non-zero acceleration; constant velocity means zero acceleration.