## Table of Contents

## Acceleration

**Acceleration** of an object is the rate of change of velocity with respect to time.

$$\begin{aligned} a &= \frac{\text{change in velocity}}{\text{time taken}} \\ &= \frac{\Delta v}{\Delta t} \\ &= \frac{v \, – u}{t_{v} – t_{u}} \end{aligned}$$

, where

- v is final velocity,
- u is initial velocity,
- $t_{v}$ is time at which object is at final velocity, $v$,
- $t_{u}$ is time at which object is at initial velocity, $u$

Details of acceleration:

- SI unit is metre per second square or metre per second per second ($m \, s^{-2}$)
- Vector quantity
**Positive Acceleration:**Occurs when an object speeds up in the direction of motion.**Negative Acceleration (Deceleration):**Occurs when an object slows down, having an opposite direction to the motion.

If the velocity of an object **increases**, the object is undergoing **acceleration**. Hence, if the velocity of an object **decreases**, it is undergoing **deceleration**.

If the velocity of the object is **constant**, the acceleration is **zero**.

An object can be stated to be accelerating if ONE of the following criteria is fulfilled:

- $\Delta$ in magnitude only
- $\Delta$ in direction only
- $\Delta$ in both magnitude and direction

An object is said to be undergoing uniform acceleration when there is a constant change in velocity per unit time.

**Note:** If the direction of an object changes, it is undergoing acceleration by definition. Change direction = change in velocity = acceleration (Might not be that important in O level)

### Average Acceleration

$$a \, = \, \frac{v-u}{t}$$

$$a \, = \, \frac{\Delta v}{\Delta t}$$

### Instantaneous Acceleration

$$\frac{dv}{dt}$$

### Uniform Acceleration

When there is a constant rate of change (increase or decrease) in the object’s velocity per unit of time, we say that the object is experiencing constant or uniform acceleration.

Time (s) | Velocity ($m s^{-1}$) | Velocity ($m s^{-1}$) |
---|---|---|

0 | 0 | 50 |

1 | 10 | 40 |

2 | 20 | 30 |

3 | 30 | 20 |

4 | 40 | 10 |

5 | 50 | 0 |

### Non-uniform Acceleration

An object experiences non-uniform acceleration when there is a varying rate of change in its velocity per unit of time.

Time (s) | Velocity ($m s^{-1}$) | Change in Velocity ($m s^{-1}$) |
---|---|---|

0 | 0 | 0 |

1 | 10 | 10 |

2 | 30 | 20 |

3 | 20 | -10 |

4 | 50 | 30 |

5 | 70 | 20 |

## Relationship Between Acceleration & Direction of Motion

#### Speeding Up Object:

- If an object is speeding up, its acceleration has the same sign as the direction of motion.
- Example: A car moving forward with positive acceleration.

#### Slowing Down Object:

- If an object is slowing down, its acceleration has the opposite sign as the direction of motion.
- Example: A car moving backward with negative acceleration.

The sign convention can be incorporated into the equations of motion.

- Positive values for displacement and acceleration indicate motion in the positive direction.
- Negative values for displacement and acceleration indicate motion in the negative direction.

## Worked Examples

### Example 1

A bus starts from rest and achieves a velocity of $20 \text{ m s}^{-1}$ in 10 seconds while moving to the right. Calculate its average acceleration.

**Click here to show/hide answer**

Average acceleration:

$$\begin{aligned} \left< a \right> &= \frac{v-u}{t} \\ &= \frac{20-0}{10} \\ &= 2 \text{ m s}^{-2} \text{ towards the right} \end{aligned}$$

### Example 2

A car travelling westwards at $30 \text{ m s}^{-1}$ suddenly comes to a halt in 5 s. Find its average acceleration.

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$$\begin{aligned} \left< a \right> &= \frac{v-u}{t} \\ &= \frac{0-30}{5} \\ &=-6 \text{ m s}^{-2} \text{ towards the west} \end{aligned}$$

Note that $-6 \text{ m s}^{-2} \text{ towards the west}$ is the same as $6 \text{ m s}^{-2} \text{ towards the east}$.

### Example 3

Would you consider a car turning around a corner at a constant speed to be accelerating?

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Yes! Acceleration is defined as the rate of change of velocity.

Velocity has two components – magnitude and direction.

When the car is turning around a corner, it’s direction is constantly changing.

Since the direction is constantly changing, the velocity will be constantly changing. (i.e. The car is accelerating!)

### Example 4

Does motion always take place in the direction of acceleration?

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No! Consider a car that is moving towards the right at a velocity of $25 \text{ m s}^{-1}$. If the driver steps on the brakes and slowly brings the car to a stop, the direction of the acceleration will be towards the LEFT during the whole deceleration process. (Even though the car is still moving towards the left!)

### Example 5

Can an object be moving if its acceleration is zero?

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Yes! An object that maintains a constant velocity (that is above/below $0 \text{ m s}^{-1}$) will be moving with a zero acceleration.

### Example 6

Explain whether or not the following particles have an acceleration:

(a) a particle moving in a straight line with constant speed,

(b) a particle moving around a curve with constant speed.

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(a) no

(b) yes. In the second case, the particle is continuously changing the direction of its velocity vector.

### Example 7

a) Clarify why the steel ball reaches the ground before the ping-pong ball when both are dropped from a 5-meter height.

b) Describe the expected velocity-time graphs for both the steel ball and the ping-pong ball during their descent, highlighting the differences in their behaviors.

**Click here to show/hide answer**

a) The steel ball reaches the ground first because it experiences less air resistance than the ping-pong ball. Due to its higher mass and density, the steel ball is less affected by air resistance, allowing it to accelerate more rapidly and reach the ground sooner.

b) For the steel ball, the velocity-time graph would show a steeper, more consistent increase in velocity due to its greater mass and minimal air resistance. On the other hand, the ping-pong ball’s graph would display a slower and less consistent increase in velocity due to its lower mass and increased susceptibility to air resistance.