Question

For sport, a 12 kgarmadillo runs onto a large pond of level, frictionless ice. The armadillo’s initial velocity is 5.0 m/salong the positive direction of an xaxis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of 17 Nin the positive direction of the yaxis. In unit-vector notation, what are the animal’s (a) velocity and (b) position vectors when it has slid for 3.0 s?

Short answer

1. Velocity after 3 is$\stackrel{r}{v}=5\hat{i}+4.3\hat{j}$
2. Position after 3 is$\stackrel{\mathrm{\Gamma }}{r}=15\hat{i}+6.4\hat{j}$

Step-by-step solution

Given

1. m = 12 kg
2. $V_x=5m/s$
   
3. $F_y=17N$
   
4. $t=3sec$
   

Understanding the concept

Acceleration of an object produced by a net force is directly proportional to the product of mass and acceleration is known as Newton’s second law of motion. So to calculate net force use the mass of the object and acceleration of the object.

Formula

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{a}$

Here,${\mathit{F}}_{\mathbf{net}}$ is the force, m is the mass of the object, and a is the acceleration of the object.

Calculate the velocity vector

(a)

To find the velocity vector, first, find velocity components along x and y directions.

We have,

$V_x=5\hat{i}$

To find y component of velocity, first calculate acceleration in y direction:

role="math" localid="1660910282733" $$\begin{gathered} F_y=m\times a_y \\ 17=a_y\times 12 \\ {a}_y=1.41m/s \end{gathered}$$

Now velocity along direction,

$\begin{array}{l}{v}_y=v_0+a_y\times t\\ v_y=1.41\times 3\\ =4.23\mathrm{m}/\mathrm{s}\end{array}$

So, velocity vector,

$$\begin{gathered} \stackrel{r}{v}=v_x\hat{i}+y_y\hat{j} \\ \stackrel{r}{v}=5\hat{i}+4.3\hat{j} \end{gathered}$$

Therefore, the position vector is $5\hat{i}+4.3\hat{j}$

Calculate the position vector after 3 s

(b)

Position along x axis:

$r_x=v_x\times t=5\times 3=15m$

Position along y axis:

$\begin{array}{l}{r}_y=V_{0}t+\frac{1}{2}{a}_y{t}^2\\ =\frac{1}{2}\times 1.41\times 3^2\\ =6.345m\end{array}$

So, position vector:

$\begin{array}{l}r=r_x\stackrel{⏜}{i}+r_y\hat{j}\\ =15\hat{i}+6.4\hat{j}\end{array}$

Therefore, the position vector is $15\hat{i}+6.4\hat{j}$.