Question

A particle moves along a straight line with the equation of motion\(s = f\left( t \right)\), where s is measured in meters and t in seconds.Find the velocity and speed when\(t = {\bf{4}}\).

\(f\left( t \right) = {\bf{10}} + \frac{{{\bf{45}}}}{{t + {\bf{1}}}}\)

Short answer

The velocity of the particle is \( - \frac{9}{5}\;{\rm{m/s}}\).

The speed of the particle is \(\frac{9}{5}\;{\rm{m/s}}\).

Step-by-step solution

Step 1:Find the velocity of the particle at \(t = {\bf{4}}\)

The velocity of the particle \(v\left( 4 \right)\) can be calculated as:

\(\begin{aligned}v\left( 4 \right) &= f'\left( 4 \right)\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {4 + h} \right) - f\left( 4 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {10 + \frac{{45}}{{\left( {4 + h} \right) + 1}}} \right) - \left( {10 + \frac{{45}}{{4 + 1}}} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{45\left( {\frac{1}{{5 + h}} - \frac{1}{5}} \right)}}{h}\\ &= - \mathop {\lim }\limits_{h \to 0} 45\left( {\frac{h}{{5\left( {5 + h} \right)h}}} \right)\\ &= - \frac{9}{5}\end{aligned}\)

So, the velocity of the particle is \( - \frac{9}{5}\;{\rm{m/s}}\).

Find the speed of the particle at \(t = {\bf{4}}\)

Therefore, the speed of the particle at \(t = 4\) is \(\left| { - \frac{9}{5}\;\;{\rm{m/s}}} \right|\), i.e.\(\frac{9}{5}\;{\rm{m/s}}\).