Question

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion \(s = \frac{1}{{{t^2}}}\), where \(t\) is measured in seconds. Find the velocity of the particle at times \(t = a,\,\,t = 1,\,\,t = 2\,\,and\,\,t = 3\).

Short answer

The velocity of the particle at times \(t = a\) is \(\frac{{ - 2}}{{{a^3}}}{\rm{ m/s}}\), at \(t = 1\) is \( - 2{\rm{ m/s}}\), at \(t = 2\) is \( - \frac{1}{4}\;{\rm{m/s}}\), at \(t = 3\) is \( - \frac{2}{{27}}\;{\rm{m/s}}\).

Step-by-step solution

Use the definition of instantaneous velocity

Apply the definition of instantaneous velocity which states that the position function \(f\left( t \right)\) at time \(t = a\) is represented as \(v\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\).

Find the velocity of the particle at times\(t = a\)

Now find the velocity by using instantaneous velocity at time \(t = a\).

\(\begin{aligned}v\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{s\left( {a + h} \right) - S\left( a \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{\left( {a + h} \right)}^2}}} - \frac{1}{{{a^2}}}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{a^2} - \left( {{a^2} + 2ah + {h^2}} \right)}}{{h{a^2}{{\left( {a + h} \right)}^2}}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{ - \left( {2a + h} \right)}}{{{{\left( {a + h} \right)}^2}}}\\ &= \frac{{ - 2a}}{{{a^2} \cdot {a^2}}}\\ &= \frac{{ - 2}}{{{a^3}}}\end{aligned}\)

Thus, the velocity of the particle at \(t = a\) is \(\frac{{ - 2}}{{{a^3}}}{\rm{ m/s}}\).

Find the velocity of the particle at \(t = 1\)

Put\(a = 1\)into\(v\left( a \right) = \frac{{ - 2}}{{{a^3}}}\).

\(\begin{aligned}v\left( 1 \right) &= \frac{{ - 2}}{{{{\left( 1 \right)}^3}}}\\ &= - 2\;{\rm{m/s}}\end{aligned}\)

Thus, the velocity of the particle at \(t = 1\) is \( - 2{\rm{ m/s}}\).

Find the velocity of the particle at times \(t = 2\)

Put\(a = 2\)into\(v\left( a \right) = \frac{{ - 2}}{{{a^3}}}\).

\(\begin{aligned}v\left( 1 \right) &= \frac{{ - 2}}{{{{\left( 2 \right)}^3}}}\\ &= - \frac{1}{4}\;{\rm{m/s}}\end{aligned}\)

Thus, the velocity of the particle at \(t = 2\) is \( - \frac{1}{4}\;{\rm{m/s}}\).

Find the velocity of the particle at times \(t = 3\)

Put\(a = 3\)into\(v\left( a \right) = \frac{{ - 2}}{{{a^3}}}\).

\(\begin{aligned}v\left( 1 \right) &= \frac{{ - 2}}{{{{\left( 3 \right)}^3}}}\\ &= - \frac{2}{{27}}\;{\rm{m/s}}\end{aligned}\)

Thus, the velocity of the particle at \(t = 3\) is \( - \frac{2}{{27}}\;{\rm{m/s}}\).