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\(t = {\bf{4}}\).

\(f\left( t \right) = {\bf{80}}t - {\bf{6}}{t^{\bf{2}}}\)

Short answer

The velocity of the particle is 32 m/s.

The speed of the particle is 32 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:

\(\begin{aligned}v\left( 4 \right) &= f'\left( 4 \right)\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {4 + h} \right) - f\left( 4 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {80\left( {4 + h} \right) - 6{{\left( {4 + h} \right)}^2}} \right) - \left( {80\left( 4 \right) - 6{{\left( 4 \right)}^2}} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{320 + 80h - 96 - 48h - 6{h^2} - 320 + 96}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{32h - 6{h^2}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {32 - 6h} \right)\\ &= 32\end{aligned}\)

So, the velocity of the particle is \(32\;{\rm{m/s}}\).

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

The speed of the particle at a particular instant represents the magnitude of the velocity at that instant.

\(\left| {v\left( 4 \right)} \right| = 32\)

Therefore, the speed of the particle at \(t = 4\) is 32 m/s.