Question

A rocket powered hockey puck moves on a horizontal frictionless table. The figure shows graphs of Vx and Vy the x and y-components of the puck's velocity. The puck starts at the orgin.

In what direction is the Puck moving at t =2s.

How far from the origin is the puck at t=5s.

Answer 1

Concept and reason

Use the concept of vectors and equations of motion to solve this problem.

First, identify the horizontal and vertical components if the velocity by using the given graphs.

Then, using the formula that relates the tangent of angle, vertical component of velocity, and horizontal component of velocity to find the direction in which the puck is moving.

Fundamentals

Displacement is the smallest distance between the two points.

The velocity is defined as the rate of change of displacement.

The vector representation of velocity is,

$V = {V_x}\hat i + {V_y}\hat j$

Here, ${V_x}$ is the horizontal component and ${V_y}$ is the vertical component, $\hat i$and $\hat j$ are the unit vectors along the x and y directions respectively.

The direction of a vector $V = {V_x}\hat i + {V_y}\hat j$is given by,

$\theta = {\tan ^{ - 1}}\left( {\frac{{{V_y}}}{{{V_x}}}} \right)$

The magnitude of a velocity vector $V = {V_x}\hat i + {V_y}\hat j$ is given by,

$\begin{array}{c}\\{\left| {\vec V} \right|^2} = V_x^2 + V_y^2\\\\V = \sqrt {V_x^2 + V_y^2} \\\end{array}$

Here, ${V_x}$ is the horizontal component and ${V_y}$ is the vertical component of the velocity.

The displacement of an object is given by,

$s = ut + \frac{1}{2}a{t^2}$

Here, u is the initial velocity, a is acceleration, and t is the time.

From the graph of x component of velocity versus time graph provided in the question, the x component of velocity at time 2 s is 16 cm/s.

From the graph of y component of velocity versus time graph provided in the question, the y component of velocity at time 2 s is 30 cm/s.

The direction of a vector $V = {V_x}\hat i + {V_y}\hat j$is given by,

$\theta = {\tan ^{ - 1}}\left( {\frac{{{V_y}}}{{{V_x}}}} \right)$

Here, ${V_x}$ is the horizontal component and ${V_y}$ is the vertical component.

Substitute $30{\rm{ cm/s}}$ for ${V_y}$ and $16{\rm{ cm/s}}$ for${V_x}$.

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{30{\rm{ cm/s}}}}{{16{\rm{ cm/s}}}}} \right)\\\\ = 61.9^\circ \\\end{array}$

The magnitude of a vector $V = {V_x}\hat i + {V_y}\hat j$ is given by,

$V = \sqrt {V_x^2 + V_y^2}$

Substitute $30{\rm{ cm/s}}$ for ${V_y}$ and $40{\rm{ cm/s}}$ for${V_x}$.

$\begin{array}{c}\\V = \sqrt {{{\left( {40{\rm{ cm/s}}} \right)}^2} + {{\left( {30{\rm{ cm/s}}} \right)}^2}} \\\\ = 50{\rm{ cm/s}}\\\end{array}$

The horizontal acceleration of the puck is,

${a_x} = \frac{{{V_x}}}{t}$

Substitute $40{\rm{ cm/s}}$ for ${V_x}$ and $5{\rm{ s}}$ for t.

$\begin{array}{c}\\{a_x} = \frac{{40{\rm{ cm/s}}}}{{5{\rm{ s}}}}\\\\ = 8{\rm{ cm/}}{{\rm{s}}^2}\\\end{array}$

The horizontal displacement is,

${s_x} = {u_x}t + \frac{1}{2}{a_x}{t^2}$

From the graph, it is clear that the x component of velocity at time t = 0 is zero.

Substitute 0 for${u_x}$, $5{\rm{ s}}$ for t, and $8{\rm{ cm/}}{{\rm{s}}^2}$ for${a_x}$.

$\begin{array}{c}\\{s_x} = \left( 0 \right)\left( {5{\rm{ s}}} \right) + \frac{1}{2}\left( {8{\rm{ cm/}}{{\rm{s}}^2}} \right){\left( {5{\rm{ s}}} \right)^2}\\\\ = 100{\rm{ cm}}\\\end{array}$

The vertical displacement is,

${s_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}$

From the graph, it is clear that the y component of velocity at time t = 0 is 30 cm/s.

Substitute $30{\rm{ cm/s}}$ for ${V_y}$ and $5{\rm{ s}}$ for t.

$\begin{array}{c}\\{s_y} = 30{\rm{ cm/s}}\left( {5{\rm{ s}}} \right)\\\\ = 150{\rm{ cm}}\\\end{array}$

The displacement of the puck is,

$s = \sqrt {s_x^2 + s_y^2}$

Substitute $100{\rm{ cm}}$ for ${s_x}$ and $150{\rm{ cm}}$ for${s_y}$.

$\begin{array}{c}\\s = \sqrt {{{\left( {100{\rm{ cm}}} \right)}^2} + {{\left( {150{\rm{ cm}}} \right)}^2}} \\\\ = 180.277{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\\\\ = 1.80{\rm{ m}}\\\end{array}$

Ans:

The puck moves at an angle $61.9^\circ$ above the horizontal.

The puck is at a distance of $1.80{\rm{ m}}$ from the origin.