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:

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

• 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 \(s=\int v\, dt\)

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 \(O\)

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.

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:

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

• 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 \(s=\int v\, dt\)

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 \(O\)

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.