Question

question 1
a particle is moving in the x - y plane, and its position vector as a function of time is:
\\( \vec { r } ( t ) = \left( a t ^ { 3 } + b \
ight) \hat { i } + c t \hat { j } \\)
with:
\\( a = 2 \frac { m } { s ^ { 3 } } , b = 7 m , c = 6 \frac { m } { s } \\)
find the speed of the particle in meters per second at time \\( t = 1 s \\).
provide at least one decimal place

Explanation:

Step1: Find the velocity vector

The velocity vector \(\vec{v}(t)\) is the derivative of the position vector \(\vec{r}(t)\).
For the \(x\)-component: \(\frac{d}{dt}(At^{3}+B)=3At^{2}\)
For the \(y\)-component: \(\frac{d}{dt}(Ct)=C\)
So \(\vec{v}(t)=3At^{2}\hat{i}+C\hat{j}\)

Step2: Substitute the values of \(A\), \(C\) and \(t\)

Given \(A = 2\frac{m}{s^{3}}\), \(C = 6\frac{m}{s}\) and \(t = 1s\)
The \(x\)-component of velocity \(v_{x}=3\times2\times(1)^{2}=6\frac{m}{s}\)
The \(y\)-component of velocity \(v_{y}=6\frac{m}{s}\)

Step3: Calculate the speed

The speed \(v\) is given by \(v=\sqrt{v_{x}^{2}+v_{y}^{2}}\)
\(v=\sqrt{6^{2}+6^{2}}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\approx8.5\frac{m}{s}\)

Answer:

\(8.5\)