Question

A snowmobile is originally at the point with position vector$\mathbf{29}.\mathbf{0}\mathbf{m}$at$\mathbf{95}.\mathbf{0}°$counterclockwise fromtheaxis, moving with velocity$\mathbf{4}.\mathbf{50}\mathbf{m}/\mathbf{s}$at$\mathbf{40}.\mathbf{0}°$. It moves with constant acceleration$\mathbf{1}.\mathbf{90}\mathbf{m}/{\mathbf{s}}^2$at. Afterhave elapsed, find

(a) its velocity and (b) its position vector

Short answer

1. The velocity of the particle after time is$-5.475\text{m}/\text{s}\hat{\text{i}}-0.35\text{m}/\text{s}\hat{\text{j}}$ .
2. The position vector of the particle is$-7.79\text{m}\hat{\mathbf{i}}+35.26\text{m}\hat{\mathbf{j}}$

Step-by-step solution

Expressions that used for solution.

• The position vector of the moving particle can be written as,$\overrightarrow{r}=x\hat{i}+y\hat{j}$
• Here,$\overrightarrow{r}$is the position vector, x is the position vector ofco-ordinate and y is the position vector of y- co-ordinate.
• The velocity of the particle can be written as,
• $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$
  
• Here,$\overrightarrow{v}$is the velocity of the particle, $v_s$is the velocity of the x-co-ordinate of the particle andis the velocity of theco-ordinate.
• The acceleration of the particle can be written as,
• $\overrightarrow{a}=a_s\hat{\mathbf{i}}+a_y\hat{\mathbf{j}}$
  
• Here,$\overrightarrow{a}$ is the acceleration of the particle, $a_x$is the acceleration of the x- co-ordinate of the particle and $a_y$is the acceleration of the y-co-ordinate.

Expression to find the initial velocity. 

The relationship between velocity and distance at constant acceleration can be written as,

$x_i=v_{m}f+\frac{1}{2}a{t}^2\text{and}{y}_i=v_{y}f+\frac{1}{2}a{t}^2$

Here, $v_{as}$is the initial velocity of the particle in x-co-ordinate and is the initial velocity in y- co-ordinate.

The relationship between velocity and time can be written as,

$v_x=v_{si}+a_{x}t$And $v_y=v_{\mu i}+a_{f}t$

The initial position vector of the particle can be calculated as,

Determine the initial velocity 

The initial velocity vector can be calculated as,

To find the velocity when it t=5.0s .

(a)

The velocity aftertime t can be calculated as,

${\overrightarrow{v}}_f=\left(v_{xi}+a_{x}t\right)\hat{\mathbf{i}}+\left(v_{yt}+a_{y}t\right)\hat{\mathbf{j}}$

Here, ${\overrightarrow{v}}_f$is the final velocity.

Substitute the value, $3.45\text{m}/\text{s}$for $v_{\text{s}}=2.90\text{m}/\text{s}$ for $v_{{\mu }^{+}}-1.785\text{m}/{\text{s}}^2$for$a_{s^{+}}-0.65\text{m}/{\text{s}}^2$for$a_y$and 5.0s for z in the above expression,

role="math" localid="1663759773553" $$\begin{gathered} {\overrightarrow{v}}_f=\left(3.45\text{m}/\text{s}+\left(-1.785\text{m}/{\text{s}}^2\right)(5.0\text{s})\right)\hat{\mathbf{i}}+\left(2.90\text{m}/\text{s}+\left(-0.65\text{m}/{\text{s}}^2\right)(5.0\text{s})\right)\hat{\mathbf{j}} \\ =-5.475\text{m}/\text{s}\hat{\text{i}}-0.35\text{m}/\text{s}\hat{\mathbf{j}} \end{gathered}$$

Hence, the velocity of the particle after time 5.0s is$-5.475\text{m}/\text{s}\hat{\text{i}}-0.35\text{m}/\text{s}\hat{\text{j}}$ .

To find its position vector.

Hence, the position vector of the particle is $-7.79\text{m}\hat{\mathbf{i}}+35.26\text{m}\hat{\mathbf{j}}.$