| 8.1 |
Displacement, Velocity and Acceleration as a Function of Time |
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| Displacement and Distance |
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Displacement, \(s\) of a particle from a fixed point is the distance of the particle from the fixed point measured on a certain direction.
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Distance is a scalar quantity that refers to the total path travelled by an object.
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| Instantaneous Displacement |
The displacement of a particle at a certain time.
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If \(O\) is a fixed point and the movement of a particle to the right is positive, then
- The displacement is negative, \(s\lt0\), meaning the particle is on the left of point \(O\).
- The displacement is zero, \(s=0\), meaning the particle is on the point \(O\).
- The displacement is positive, \(s\gt 0\), meaning the particle is on the right of point \(O\).
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| Velocity and Speed |
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A velocity, \(v\) is a vector quantity which defined as the rate of change of displacement with time.
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Speed is a scalar quantity which defined as the rate of change of distance with time.
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The velocity of an object at a certain time.
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If \(O\) is a fixed point and the movement of a particle to the right is positive, then
- The velocity is positive, \(v\gt 0\), meaning that the particle moves to the right.
- The velocity is zero, \(v=0\), meaning that the particle is at rest, that is, the particle is stationary.
- The velocity is negative, \(v\lt 0\), meaning that the particle moves to the left.
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| Acceleration |
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Acceleration, \(a\) is the rate of change of velocity with time. The acceleration function, \(a\) is a function of time, \(a=f(t)\) and is a vector quantity that has magnitude and direction.
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The rate of change of velocity with time of an object that moves is the same at any time.
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The rate of change of velocity with time of an object that moves is different at any time.
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| Instantaneous Acceleration |
An acceleration, \(a\) at a certain time, \(t\) and can be obtained by determining the gradient of tangent of velocity-time graph at time, \(t\).
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If the movement of a particle to the right is positive, then
- The acceleration is positive, \(a\gt 0\), meaning the velocity of particle is increasing with time.
- The acceleration is zero, \(a=0\), meaning the velocity of particle is either maximum or minimum.
- The acceleration is negative, \(a\lt 0\), meaning the velocity of particle is decreasing with time.
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| Example |
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A particle moves along a straight line and passes through a fixed point \(O\).
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Its displacement, \(s\) m, of the particle \(t\) seconds after it starts moving is given as |
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\(s=4+8t-t^2\). |
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Calculate the instantaneous displacement, in m, and determine the position of the particle from point \(O\) when \(t=10\). |
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| (b) |
Its velocity, \(v\) ms\(^{-1}\), at \(t\) after passing through the point \(O\) is given by |
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\(v=3t-12\). |
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Calculate the initial velocity in ms\(^{-1}\), of the particle. |
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| (c) |
At \(t\) seconds after passing through \(O\), its acceleration, \(a\) ms\(^{-2}\), is given by |
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\(a=12-4t\). |
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Calculate the instantaneous acceleration, in ms\(^{-2}\), of the particle when \(t=7\). |
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(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 \(16\) 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 \(-12\) 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 \(-16\) ms\(^{-2}\).
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