| |
- Displacement is a vector quantity that has a magnitude and a direction
- Distance scalar quantity
- The value of displacement can be positive, zero or negative
|
| |
| Displacement, \(s\) |
| |
|
|
| |
A particle from a fixed point is the distance of the particle from the fixed point measured on a certain direction.
|
|
| |
|
|
|
|
| |
| Distance |
| |
|
|
| |
A scalar quantity that refers to the total path travelled by an object.
|
|
| |
|
|
|
|
| |
| |
|
|
| |
If \(O\) is a fixed point and the movement of a particle to the right is positive, then |
|
| |
|
|
| |
| 1. |
The displacement is negative, \(s \text{ < } 0\), meaning the particle is on the left of point \(O\). |
| |
|
| 2. |
The displacement is zero, \(s=0\), meaning the particle is on the point \(O\). |
| |
|
| 3. |
The displacement is positive, \(s \text{ > } 0\), meaning the particle is on the right of point \(O\). |
|
|
| |
|
|
|
| |
- Velocity, \(v\) is a vector quantity that has a magnitude and a direction
- Speed is scalar quantity
|
| |
| Velocity, \(v\) |
| |
|
|
| |
The rate of change of displacement with time. |
|
| |
|
|
|
| Speed |
| |
|
|
| |
The rate of change of distance with time. |
|
| |
|
|
|
|
| |
- The velocity of an object at a certain time is called an instantaneous velocity
- In general,
| |
|
|
| |
If \(O\) is a fixed point and the movement of a particle to the right is positive, then |
|
| |
|
|
| |
| 1. |
The velocity is positive, \(v\text{ > }0\), meaning that the particle moves to the right.
|
| |
|
| 2. |
The velocity is zero, \(v=0\), meaning that the particle is at rest, that is, the particle is stationary. |
| |
|
| 3. |
The velocity is negative, \(v\text{ < }0\), meaning that the particle moves to the left. |
|
|
| |
|
|
- Acceleration is a vector quantity that has magnitude and direction
- The acceleration function is a function of time
|
| |
| Acceleration, \(a\) |
| |
|
|
| |
Acceleration is the rate of change of velocity with time.
|
|
| |
|
|
|
| Uniform Acceleration |
Non-uniform Acceleration |
| |
|
|
| |
If the rate of change of velocity with time of an object that moves is the same at any time
|
|
| |
|
|
|
| |
|
|
| |
If the rate of change of velocity with time is different at any time
|
|
| |
|
|
|
- An acceleration, \(a\) at a certain time, \(t\) is called an instantaneous acceleration
- It can be obtained by determining the gradient of tangent of velocity-time graph at time, \(t\)
- In general,
|
| |
|
|
| |
If the movement of a particle to the right is positive, then
|
|
| |
|
|
| |
| 1. |
The acceleration is positive, \(a\text{ > }0\), meaning the velocity of particle is increasing with time.
|
| |
|
| 2. |
The acceleration is zero, \(a=0\), meaning the velocity of particle is either maximum or minimum. |
| |
|
| 3. |
The acceleration is negative,  meaning the velocity of particle is decreasing with time
|
|
|
| |
|
|
|
| |
| Example |
| |
|
|
| |
A particle moves along a straight line and passes through a fixed point \(O\).
|
|
| |
| (a) |
Its displacement, \(s \text{ m}\), of the particle \(t\) seconds after it starts moving is given as
\(s=4+8t-t^2\).
Calculate the instantaneous displacement, in \(\text{m}\), and determine the position of the particle from point \(O\) when \(t=10\).
|
| |
|
| (b) |
Its velocity, \(v\text{ ms}^{-1}\), at \(t\) after passing through the point \(O\) is given by
\(v=3t-12\).
Calculate the initial velocity in \(\text{ms}^{-1}\), of the particle.
|
| |
|
| (c) |
At \(t\) seconds after passing through \(O\), its acceleration, \(a \text{ ms}^{-2}\), is given by
\(a=12-4t\).
Calculate the instantaneous acceleration, in \(\text{ms}^{-2}\), of the particle when \(t=7\).
|
| |
|
| |
|
| Solution: |
| |
|
| (a) |
Given \(s=4+8t-t^2\).
When \(t=10\),
\(s=4+8(10)-(10)^2\\ s=4+80-100\\ s=-16\)
Therefore, the particle is located \(16\text{ m}\) from the fixed point \(O\) when \(t=10\).
|
| |
|
| (b) |
Given \(v=3t-12\).
At initial velocity, \(t=0\),
\(v=3(0)-12\\ v=-12\)
Hence, the initial velocity of the particle is \(-12\text{ ms}^{-1}\).
|
| |
|
| (c) |
Given \(a=12-4t\).
When \(t=7\),
\(a=12-4(7)\\ a=-16\)
Therefore, the instantaneous acceleration of the particle when \(t=7\) is \(-16\text{ ms}^{-2}\)
|
| |
|
|
|
|
|
|
|
