Question

A frightened rabbit moving at 6.0 m/sdue east runs onto a large area of level ice of negligible friction. As the rabbit slides across the ice, the force of the wind causes it to have a constant acceleration of$\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, due north. Choose a coordinate system with the origin at the rabbit’s initial position on the ice and the positive xaxis directed toward the east. In unit-vector notation, what are the rabbit’s (a) velocity and (b) position when it has slide for 3.00 s?

Short answer

a) The velocity of the rabbit after 3.00 s is$\left(6.0\stackrel{⏜}{i}+4.2\stackrel{⏜}{j}\right)\mathrm{m}/\mathrm{s}$

b) The position of the rabbit after 3.00 s is$\left(18\stackrel{⏜}{i}+6.3\stackrel{⏜}{j}\right)\mathrm{m}/\mathrm{s}$

Step-by-step solution

The given data

• Initial velocity of rabbit in the x-direction,$V_{ix}=6.00\mathrm{m}/\mathrm{s}$.
• Constant acceleration of rabbit in the y-direction,$a_y=1.4\mathrm{m}/{\mathrm{s}}^2$.
• Time of slide,t=3.00 s .

Understanding the concept of kinematic equation

The kinematic equations relate the initial velocity, final velocity, acceleration, displacement, and time. The combination of these equations can be used to find out the unknown quantities in the problem.

We use x and y components to find vectors of velocity and position. We can find the velocity of a rabbit after 3.00s by finding the velocity along the x and y direction using a kinematic equation; similarly, we can find the position using a kinematic equation.

Formulae:

The first kinematic equation of motion,$V_f=V_i+at$ (i)

The second kinematic equation of motion,$d=V_{i}t+\left(\frac{1}{2}\right)a{t}^2$ (ii)

Here, $V_i$is the initial velocity,$V_i$ is final velocity, is acceleration,t is time and d is displacement.

a) Calculation for the velocity of rabbit after 3.00s

Components of initial velocity are $V_{ix}=6.00\mathrm{m}/\mathrm{s},{\mathrm{V}}_{\mathrm{iy}}=0$. The final x component will be the same because acceleration is along the y-direction.

Using equation (i) we get the final velocity as:

$V_f=v_i+at$

Substitute the given values in the equation (i).

$$\begin{gathered} V_{fy}=0+\left(1.4\mathrm{m}/{\mathrm{s}}^2\right)\left(3.00\mathrm{s}\right) \\ =4.2\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the final velocity of the rabbit after 3.00s is given as:

$V_f=V_{fx}\stackrel{⏜}{i}+V_{fy}\stackrel{⏜}{j}$

Substitute the values in the above equation.

role="math" localid="1657022159573" $V_f=\left(6.0\mathrm{m}/\mathrm{s}\stackrel{⏜}{i}+4.2\mathrm{m}/\mathrm{s}\stackrel{⏜}{j}\right)$

Hence, the velocity of the rabbit is$\left(6.0\stackrel{⏜}{i}+4.2\stackrel{⏜}{j}\right)\mathrm{m}/\mathrm{s}$ .

b) Calculation for the position of rabbit after 3.00s

The x-position of the rabbit can be calculated using equation (ii).

$d=v_{i}t+\left(\frac{1}{2}\right)a{t}^2$

Substitute the given values in the above equation.

$$\begin{gathered} d_x=\left(6.00\mathrm{m}/\mathrm{s}\right)\left(3.00\mathrm{s}\right)+0 \\ =18.0\mathrm{m} \end{gathered}$$

Substitute the given values in equation (ii).

$$\begin{gathered} d_y=0+\left(\frac{1}{2}\right)\left(1.4\mathrm{m}/{\mathrm{s}}^2\right){\left(3.00\mathrm{s}\right)}^2 \\ =6.3\mathrm{m} \end{gathered}$$

Final position of rabbit can be given through the resultant value, that is:

$d=d_x\stackrel{⏜}{i}+d_y\stackrel{⏜}{j}$

Substitute the values in the above equation.

$d=\left(18\mathrm{m}\stackrel{⏜}{i}+6.3\mathrm{m}\stackrel{⏜}{j}\right)$

Hence, the position of the rabbit is$\left(18\stackrel{⏜}{i}+6.3\stackrel{⏜}{j}\right)\mathrm{m}$ .