Integration in Kinematics of Linear Motion

8.3 Integration in Kinematics of Linear Motion
 
The image illustrates the relationship between acceleration function and velocity function. It shows that acceleration (a) is represented as h(t), which leads to velocity (v) being the integral of acceleration over time (∫a dt), and finally, velocity (v) is represented as g(t). The image has a blue background with white text and includes the Pandai logo at the bottom.
 
Determining the Instantaneous Velocity of a Particle from its Acceleration Function

When the acceleration function \(a=\dfrac{dv}{dt}\), the velocity function, \(v\) can be determined by integrating the acceleration function, \(a\) with respect to time, \(t\) which is:

\(v=\int a\, dt\)

 
Determining the instantaneous Displacement of a Particle from its Velocity and Acceleration Functions
  • When the velocity function, \(v\) is given as a function of time, \(t\), the displacement function, \(s\) can be obtained by performing an integration, which is:

\(s=\int v\,dt\)

  • When the acceleration function, \(a\) is given as a function of time, \(t\), the displacement function, \(s\) can be obtained by performing two consecutive integrations, which are:

\(v=\int a \, dt\) and \(s=\int v\, dt\)

 
Example
Question

A particle moves along a straight line and passes through a fixed point \(O\) with a velocity of \(12\) ms\(^{-1}\). The acceleration, \(a\) ms\(^{-2}\), at \(t\) seconds after passing through \(O\) is given by \(a=4-2t\).

Determine the instantaneous displacement, in m, of the particle from \(O\)

(a) when \(t=3\),
(b) when the particle is at rest.
Solution

Velocity function, \(v\) is given by:

\(\begin{aligned} v&=\int a\, dt \\ &=\int (4-2t)\, dt \\ &=4t-t^2+c. \end{aligned}\)

When \(t=0\) and \(v=12\), then:

\(\begin{aligned} 12&=4(0)-0^2+c \\ c&=12. \end{aligned}\)

Hence, at time \(t\)\(v=12+4t-t^2\).


Displacement function, \(s\) is given by:

\(\begin{aligned} s&=\int v\,dt \\ &=\int (12+4t-t^2)\, dt \\ &=12t+2t^2-\dfrac{1}{3}t^3+c. \end{aligned}\)

When \(t=0\) and \(s=0\), then:

\(\begin{aligned} 0&=12(0)+2(0)^2-\dfrac{1}{3}(0)^3+c \\ c&=0. \end{aligned}\)

Hence, at time \(t\)\(s=12t+2t^2-\dfrac{1}{3}t^3\).


(a)

When \(t=3\),

\(\begin{aligned} s&=12(3)+2(3)^2-\dfrac{1}{3}(3)^3 \\ &=36+18-9 \\ &=45. \end{aligned}\)

Hence, the instantaneous displacement when \(t=3\) is \(45\) m.


(b)

When the particle is at rest, \(v=0\).

Then,

\(\begin{aligned} 12+4t-t^2&=0 \\ t^2-4t-12&=0 \\ (t+2)(t-6)&=0 \end{aligned}\)

Since \(t\ge 0\)\(t=6\).

When \(t=6\),

\(\begin{aligned} s&=12(6)+2(6)^2-\dfrac{1}{3}(6)^3 \\ &=72+72-72 \\ &=72. \end{aligned}\)

Hence, the instantaneous displacement when the particle is at rest is \(72\) m.

 

Integration in Kinematics of Linear Motion

8.3 Integration in Kinematics of Linear Motion
 
The image illustrates the relationship between acceleration function and velocity function. It shows that acceleration (a) is represented as h(t), which leads to velocity (v) being the integral of acceleration over time (∫a dt), and finally, velocity (v) is represented as g(t). The image has a blue background with white text and includes the Pandai logo at the bottom.
 
Determining the Instantaneous Velocity of a Particle from its Acceleration Function

When the acceleration function \(a=\dfrac{dv}{dt}\), the velocity function, \(v\) can be determined by integrating the acceleration function, \(a\) with respect to time, \(t\) which is:

\(v=\int a\, dt\)

 
Determining the instantaneous Displacement of a Particle from its Velocity and Acceleration Functions
  • When the velocity function, \(v\) is given as a function of time, \(t\), the displacement function, \(s\) can be obtained by performing an integration, which is:

\(s=\int v\,dt\)

  • When the acceleration function, \(a\) is given as a function of time, \(t\), the displacement function, \(s\) can be obtained by performing two consecutive integrations, which are:

\(v=\int a \, dt\) and \(s=\int v\, dt\)

 
Example
Question

A particle moves along a straight line and passes through a fixed point \(O\) with a velocity of \(12\) ms\(^{-1}\). The acceleration, \(a\) ms\(^{-2}\), at \(t\) seconds after passing through \(O\) is given by \(a=4-2t\).

Determine the instantaneous displacement, in m, of the particle from \(O\)

(a) when \(t=3\),
(b) when the particle is at rest.
Solution

Velocity function, \(v\) is given by:

\(\begin{aligned} v&=\int a\, dt \\ &=\int (4-2t)\, dt \\ &=4t-t^2+c. \end{aligned}\)

When \(t=0\) and \(v=12\), then:

\(\begin{aligned} 12&=4(0)-0^2+c \\ c&=12. \end{aligned}\)

Hence, at time \(t\)\(v=12+4t-t^2\).


Displacement function, \(s\) is given by:

\(\begin{aligned} s&=\int v\,dt \\ &=\int (12+4t-t^2)\, dt \\ &=12t+2t^2-\dfrac{1}{3}t^3+c. \end{aligned}\)

When \(t=0\) and \(s=0\), then:

\(\begin{aligned} 0&=12(0)+2(0)^2-\dfrac{1}{3}(0)^3+c \\ c&=0. \end{aligned}\)

Hence, at time \(t\)\(s=12t+2t^2-\dfrac{1}{3}t^3\).


(a)

When \(t=3\),

\(\begin{aligned} s&=12(3)+2(3)^2-\dfrac{1}{3}(3)^3 \\ &=36+18-9 \\ &=45. \end{aligned}\)

Hence, the instantaneous displacement when \(t=3\) is \(45\) m.


(b)

When the particle is at rest, \(v=0\).

Then,

\(\begin{aligned} 12+4t-t^2&=0 \\ t^2-4t-12&=0 \\ (t+2)(t-6)&=0 \end{aligned}\)

Since \(t\ge 0\)\(t=6\).

When \(t=6\),

\(\begin{aligned} s&=12(6)+2(6)^2-\dfrac{1}{3}(6)^3 \\ &=72+72-72 \\ &=72. \end{aligned}\)

Hence, the instantaneous displacement when the particle is at rest is \(72\) m.