Question

The position of a body moving along a coordinate line at time \(t\) is \(s=(4+6 t)^{3 / 2},\) with \(s\) in meters and \(t\) in seconds. Find the body's velocity and acceleration when \(t=2 \mathrm{sec} .\)

Short answer

At \(t = 2s\), the body has a velocity of \[v(2) = \frac{3}{2}(4+6\cdot2)^{1/2} \cdot 6\] m/s and an acceleration of \[a(2) = \frac{3}{4} \cdot 6 \cdot (4+6\cdot2)^{-1/2} \cdot 6\] m/s² when evaluated.

Step-by-step solution

Calculating Velocity

The velocity of the body can be obtained by differentiating the displacement function \(s=(4+6t)^{3 / 2}\). This can be achieved using the chain rule for differentiation: \[v(t) = \frac{ds}{dt} = \frac{3}{2}(4+6t)^{1/2} \cdot 6 \]

Evaluate Velocity at \(t=2s\)

Now we substitute \(t=2\) into the derivative function to find the velocity at that instant. We get: \[v(2) = \frac{3}{2}(4+6\cdot2)^{1/2} \cdot 6\]

Calculating Acceleration

The acceleration can be obtained by differentiating the velocity function, again by using the chain rule for differentiation: \[a(t) = \frac{dv}{dt} =\frac{3}{4} \cdot 6 \cdot (4+6t)^{-1/2} \cdot 6\]

Evaluate Acceleration at \(t=2s\)

Finally, substitute \(t=2\) into the acceleration function to find the acceleration at that time: \[a(2) = \frac{3}{4} \cdot 6 \cdot (4+6\cdot2)^{-1/2} \cdot 6\]