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Kinematic Questions

Questions

A particle is constrained to travel in a plane so that its position is described by
\[\vec{r}=\ \left(17\frac{m}{s^2}\right)t^2 \rm \hat i+\ \left[\left(20\frac{m}{s}\right)t\ -\ \left(6\frac{m}{s^4}\right)t^4\right]\hat{\mathrm{j}}\] 
(a) Consider the motion of the particle during the time interval from $t=0\,{\rm s}$ to $t=5\,{\rm s}$? What are the displacement, average velocity and average acceleration over this time interval.
(b) Find the position, velocity and acceleration at $t=8\,{\rm s}$.

(a) Displacement of an object as it moves from an initial position $\vec r_1$ to a final position $\vec r_2$ is defined as $\Delta \vec r=\vec r_2-\vec r_1$. So we must first find the position of the particle in the $0\,{\rm s}$ and  $5\,{\rm s}$ . substitute them in the $\vec{r}(t)$ so
\begin{align*} 
\vec{r}\left(t=5\,{\rm s}\right)&=\ \left(17\frac{m}{s^2}\right){\left(5s\right)}^2\,\rm \hat i+\ \left[\left(20\frac{m}{s}\right)\left(5s\right)-\ \left(6\frac{m}{s^4}\right){\left(5s\right)}^4\right]\hat{\mathrm{j}}\\ 
&=425\ \rm \hat{i}+\left(100-3750\right)\rm \hat{j}=425\ \rm \hat{i}-3650\ \rm \hat{j}\ \ (m)
\end{align*}
\[\vec{r}\left(t=0\,{\rm s}\right)=\ \left(17\frac{m}{s^2}\right){\left(0\,{\rm s}\right)}^2\rm \hat i+\ \left[\left(20\frac{m}{s}\right)\left(0\,{\rm s}\right)-\ \left(6\frac{m}{s^4}\right){\left(0s\right)}^4\right]\hat{\mathrm{j}}=0\] 
\[\Delta \vec{r}={\vec{r}}_2\left(t=5\,{\rm s}\right)-{\vec{r}}_1\left(t=0\right)=425\ \rm \hat{i}-3650\ \rm \hat{j}\] 
By definition, the average velocity is $\vec{v}=\mathrm{\Delta }\vec{r}/\mathrm{\Delta }t\ $ therefore
\[\vec{\overline{v}}=\frac{\Delta\vec{r}}{\Delta t}=\frac{425\ \hat{i}-3650\ \rm \hat{j}}{5-0}=85\,\rm\hat{i}-730\,\rm\hat{j}\left(\frac{m}{s}\right)\] 
Average acceleration is defined as $\vec{\overline{a}}=\Delta \vec{v}/\Delta t$ where $\vec{v}$ is the instantaneous velocity of the particle i.e. $\vec{v}=d\vec{r}/dt$
So first find the change in the instantaneous velocity of the particle during time interval $0$ to $5\mathrm{s}$ then calculate the average acceleration.
\begin{align*}
\vec{v}=\frac{d\vec{r}}{dt}&=\frac{d}{dt}\left(\left(17\frac{m}{s^2}\right)t^2\,\rm \hat i+\ \left[\left(20\frac{m}{s}\right)t\ -\ \left(6\frac{m}{s^4}\right)t^4\right]\hat{\mathrm{j}}\right)\\
&=34\,t\ \rm\hat{i}+\left[20-24t^3\right]\rm\hat{j}
\end{align*}
\begin{gather*}
\vec{v}\left(t=5\right)=34\left(5\right)\rm\hat i+\left[20-24{\left(5\right)}^3\right]\,\rm\hat j=170\,\rm\hat i-2980\,\rm\hat j\left(\frac{m}{s}\right)\\
\vec{v}\left(t=0\right)=20\,\rm\hat j\left(\frac{m}{s}\right)
\end{gather*}
\begin{align*}
\Delta\vec{v}&=\vec{v}\left(t=5\right)-\vec{v}\left(t=0\right)\\
&=170\,\rm\hat i+\left(-2980-20\right)\,\rm\hat j\\
&=170\,\rm\hat i-3000\,\rm\hat j
\end{align*}
\[\vec{\vec{a}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{170\,\rm\hat i-3000\,\rm\hat j}{5-0}=34\,\rm\hat i-600\,\rm\hat j\left(\frac{m}{s^2}\right)\] 
(b) Now we must calculate the instantaneous quantities in $t=8\mathrm{s}$
\[\vec r\left(t=8\,{\rm s}\right)=\left(17\frac{m}{s^2}\right)8^2 \rm \hat i +\ \left[\left(20\frac{m}{s}\right)8\ -\ \left(6\frac{m}{s^4}\right)8^4\right]\hat{\mathrm{j}}=1088\,\rm\hat i-24416\,\rm\hat j\] 
\[\vec v=\frac{d\vec r}{dt}=34t\,\rm\hat i+\left[20-24t^3\right]\,\rm\hat j=34\left(8\right)+\left[20-24{\left(8\right)}^3\right]\,\rm\hat j=272\,\rm\hat i-12268\,\rm\hat j\] 
\[\vec a=\frac{d\vec v}{dt}=34\,\rm\hat i-72t^2\,\rm\hat j=34\,\rm\hat i-72{\left(8\right)}^2\,\rm\hat j=34\,\rm\hat i-4608\,\rm\hat j\] 
 

The position of a particle as a function of time is given as:
\[\vec{r}\left(t\right)=R\,{\cos \omega t\ }\,\rm\hat i+R\,{\sin \omega t\ }\] 
Determine:
(a) Instantaneous velocity vector, $\vec{v}(t)$
(b) The angle between $\vec{r}\left(t\right)$ and $\vec{v}(t)$
(c) Instantaneous acceleration vector
(d) What is the mathematical relation between $\left|\vec{a}\right|$ and $\left|\vec{v}\right|$?
(e) Using this relation (in part d), comment on the nature of this motion.

(a) Using the definition of the instantaneous velocity, we have $$\vec{v}\left(t\right)=\frac{d\vec{r}}{dt}=-\hat{x}\ R\omega {\sin \omega t\ }+\hat{y}\ R\omega {\cos \omega t\ }$$
(b) The angle between two vectors $\vec{A}$ and $\vec{B}$ is defined as ${\cos \alpha \ }=\frac{\vec{A}.\vec{B}}{\left|A\right|\left|B\right|}$. first compute the inner product of them: 
\begin{align*}
\vec{r}\left(t\right).\vec{v}\left(t\right) &=\left(R{\cos \omega t\ }\hat{x}+R{\sin \omega t\ }\hat{y}\right).\left(-\hat{x}\ R\omega \,{\sin R\omega \ }+\hat{y}\ R\omega \,{\cos R\omega \ }\right)\\
& =-R^2\omega \,{\sin \omega t\ }\,{\cos \omega t\ }\left(\hat{x}.\hat{x}\right)+R^2\omega \,{\sin \omega t\ }{\cos \omega t\ }\left(\hat{y}.\hat{y}\right)=0
\end{align*} 
Thus, ${\cos \alpha \ }=0\Rightarrow \ \alpha =90{}^\circ \ or\ 270{}^\circ $
(c) $$\vec{a}=\frac{d\vec{v}}{dt}=-\hat{x}R\,{\omega }^2{\cos \omega t\ }-\hat{y}R{\omega }^2{\sin \omega t\ }$$
(d) The magnitude of a vector $\vec{a}=a_x\hat{x}+a_y\,\rm\hat j$ is $\left|\vec{a}\right|=\sqrt{a^2_x+a^2_y}$. So 
\[\left|\vec{v}\right|=\sqrt{{\left(R\omega {\sin \omega t\ }\right)}^2+{\left(R\omega {\cos \omega t\ }\right)}^2}=R\omega \] 
\[\left|\vec{a}\right|=\sqrt{{\left(R{\omega }^2{\cos \omega t\ }\right)}^2+{\left(-R{\omega }^2{\sin \omega t\ }\right)}^2}=R{\omega }^2\] 
\[\therefore \ \left|\vec{a}\right|=\frac{{\left|\vec{v}\right|}^2}{R}\] 
(e) This is the defining equation for uniform circular motion. From the beginning, one can infer the nature of the circular motion from $\vec{r}\left(t\right)$ since it defines a circle!
 

Let $\vec{A}=2\hat{x}-2\hat{y}$ and $\vec{B}=2\hat{x}+2\hat{y}$. What is the value of the angle that lies between these two vectors?

Recall that the angle between two vectors $\vec{A},\ \vec{B}$ is determined by the following relation
\[{\cos \theta \ }=\frac{\vec{A}.\vec{B}}{\left|A\right|\left|B\right|}\] 
Where dot denotes the scalar product and $\left|\right|$ is the magnitude of vector. 
\[{\cos \theta \ }=\frac{\left(2\hat{x}-2\hat{y}\right).\left(2\hat{x}+2\hat{y}\right)}{\sqrt{2^2+{\left(-2\right)}^2}\sqrt{2^2+2^2}}=\frac{4-4}{denominator}=0\Rightarrow \theta =90{}^\circ \] 
 

If two vectors are given as $\vec{A}=-12\,\rm\hat i+8\,\rm\hat j$ and $\vec{B}=24\,\rm\hat i+18\,\rm\hat j$, determine the resultant vector $\vec{R}=2\vec{A}-\frac{1}{3}\vec{B}$, and also find its magnitude ($R\ or\ \left|R\right|$) and direction (angle $\theta $ between vector and horizontal line)

The calculation of the resultant vector $R$ is straightforward as follows
\[\vec{R}=2\left(-12\,\rm\hat i+8\,\rm\hat j\right)-\frac{1}{3}\left(24\,\rm\hat i+18\,\rm\hat j\right)=-24\,\rm\hat i+16\,\rm\hat j-8\,\rm\hat i-6\,\rm\hat j\] 
\[\therefore \ \ \vec{R}=-32\,\rm\hat i+10\,\rm\hat j\] 
The magnitude of vector $\vec{A}$ with vector components $A_x$ and $A_y$ is given by 
\[\left|\vec{A}\right|=\sqrt{A^2_x+A^2_y}\] 
And its direction with respect to the horizontal line is 
\[\theta ={{\tan}^{-1} \left(\left|\frac{A_y}{A_x}\right|\right)\ }\] 
Therefore, the magnitude and direction of the resultant vector $\vec{R}$ is 
\[\left|\vec{R}\right|=\sqrt{{\left(-32\right)}^2+{\left(10\right)}^2}=33.5\mathrm{m}\] 
\[{\theta }_R={{\tan}^{-1} \left(\left|\frac{10}{-32}\right|\right)\ }=17.4{}^\circ \] 
Since $A_y>0$ and $A_x<0$ so the resultant vector $\vec{R}$ lies in the second quadrant of the coordinate system.
Thus $\vec{R}=(33.5\ \mathrm{m,\ 17.4{}^\circ \ north\ of\ west)}$ is shown in the figure below.

Find the angle between the two vectors $\vec{F}=\left(3\,\rm\hat i-4\,\rm\hat j+5\hat{k}\right)\mathrm{N}$ and $\vec{s}=\left(2\,\rm\hat i-3\hat{k}\right)\mathrm{\ N}$.

The angle between two vectors is given by the following relation
\begin{align*}
{\cos \theta \ }&=\frac{\vec{A}.\vec{B}}{\left|A\right|\left|B\right|}\\
&=\frac{\left(3\,\rm\hat i-4\,\rm\hat j+5\hat{k}\right).\left(2\,\rm\hat i-3\hat{k}\right)}{\sqrt{3^2+{\left(-4\right)}^2+5^2}\sqrt{2^2+{\left(-3\right)}^2}}\\
&=\frac{3\times 2+5\times \left(-3\right)}{\sqrt{50}\sqrt{13}}\\
&=110.67{}^\circ
\end{align*}
$\vec{A}.\vec{B}$ is scalar product which is defined as 
\begin{align*}
A.B &=\left(A_x\,\rm\hat i+A_y\,\rm\hat j+A_z\hat{k}\right).\left(B_x\,\rm\hat i+B_y\,\rm\hat j+B_z\hat{k}\right)\\
& =A_xB_x\left(\underbrace{\rm\hat i.\rm\hat i}_{1}\right)+A_xB_y\left(\underbrace{\rm\hat i.\,\rm\hat j}_{0}\right)+A_zB_x\left(\underbrace{\hat{k}.\rm\hat i}_{0}\right)+\dots
\end{align*}
\[A_xB_x+A_yB_y+A_zB_z\] 
$\left|\vec{A}\right|$ and $\left|\vec{B}\right|$ are the magnitudes of the corresponding vectors $\vec{A}$ and $\vec{B}$ which are defined as
\[\left|\vec{A}\right|=\sqrt{A^2_x+A^2_y+A^2_z}\] 
 

A bullet is fired horizontally from the top of a cliff that is $80\ \mathrm{m}$ above a big lake. If the muzzle (initial) speed of the bullet is $400\ \mathrm{m/s}$, how far from the bottom of the cliff does the bullet strike the surface of the lake? Neglect air resistance. 

A pictorial representation of the problem is shown in the figure below.


Put a coordinate at the starting point. Since the hitting point of the bullet is $80\ \mathrm{m}$ below the coordinate so the coordinate of the landing point is $(x=?,y=-80\, \mathrm{m)}$. first using the equation $y=-\frac{1}{2}gt^2+v_{0y}t+y_0$ find the time required to reaches the bullet to the ground.
\[y=-\frac{1}{2}gt^2\to t=\sqrt{\frac{2y}{-g}}=\sqrt{2\times \frac{(-80)}{(-10)}}=4\ \mathrm{s}\] 
Now using the relation between uniform velocity and displacement i.e. $x=Vt$ we obtain
\[x=Vt=\left(400\right)\left(4\right)=1600\, \mathrm{m}\ \mathrm{\to }\mathrm{x=1 mile}\] 
 

A steel ball is dropped from the top of the Leaning Tower of Pizza, which is $56\ \mathrm{m}$ high. With air resistance neglected, approximately how long does it take the ball to hit the ground?

This is a free fall problem. So use the following kinematic equation to find the desired time
\[y=-\frac{1}{2}gt^2+v_0t+y_0\] 
Where $v_0$ is the initial velocity of the ball and $y_0$ is initial height. In kinematic problems, we must first consider a reference point. Let the top of the tower be the origin of the coordinates so the hitting point has the coordinate $(x=0,y=-56\,\mathrm{m)}$. Since the ball is dropped so its initial velocity is zero ($v_0=0$).
\[y=-\frac{1}{2}gt^2+v_0t+y_0\] 
\[-56=-\frac{1}{2}\left(9.8\right)t^2+0+0\Rightarrow t=\sqrt{\frac{2\times 56}{9.8}}=3.38\ \mathrm{s}\] 
$y_0=0$ since the ball has sat on the origin. 
 

Suppose that you are climbing the Leaning Tower, and you throw your cellphone to a friend on the ground, who catches it at a point $30\ \mathrm{m}$ below the point where you released it.
You throw it straight out, with an initial horizontal velocity of $10\ \mathrm{m/s}$ (and zero initial vertical velocity). By neglecting air resistance, what is the speed of the cellphone when your friend catches it? 
Find the components of the velocity at the hitting point then use the Pythagorean theorem to determine its magnitude.

Horizontal motion is uniform with speed $v_{0x}=v_x=10\ \mathrm{m/s}$ but vertical motion is accelerating. So use the following kinematic equation to find its velocity. 
\[v^2-v^2_0=2a\left(y-y_0\right) \to V^2_{y}-v^2_{0y}=2(-g)(y-y_0)\] 
\[v^2-0=2\left(-9.8\right)\left(-30-0\right) \Rightarrow v^2=588\frac{m^2}{s^2}\] 
\[\Rightarrow v=\sqrt{v^2_x+v^2_y}=\sqrt{{10}^2+588}=26\ \mathrm{m/s}\] 

A ball is thrown vertically upward with an initial speed of $20\ \mathrm{m/s}$. Neglect air resistance and assume constant acceleration. 
(a) How long does it take for the ball to reach its maximum height?
(b) What is that maximum height?
(c) At what height is the ball when its speed is $10\ \mathrm{m/s}$?
(d) How long does it take for the ball to return to the position at which it was released?
(e) What is its speed when it returns?

(a) Given data are $v_0=20\ \mathrm{m/s}$ , $a=-g$ so use the following kinematic equation to find the desired time
\[v=v_0-gt\] 
At the maximum height, the velocity of the ball is zero i.e. $v=0$. Therefore, we obtain
\[0=20-9.8t\Rightarrow t=2.04\ \mathrm{s}\] 
(b) let the starting point as origin so $y_0=0$. Using the following kinematic relation we get
\[v^2-v^2_0=-2g\left(y-y_0\right)\] 
\[0-{20}^2=-2\left(9.8\right)\left(y-0\right)\Rightarrow y=\frac{400}{2\times 9.8}=20.4\ \mathrm{m}\] 
(c) Using the equation $v^2-v^2_0=-2g(y-y_0)$, we obtain
\[{10}^2-{20}^2=-2\left(9.8\right)\left(y-0\right)\Rightarrow y=\frac{300}{2\times 9.8}=15.3\ \mathrm{m}\] 
(d) Since there is no air resistance, we can multiply the result of part (a) by two and get $t_{tot}=2\times 2.04=4.08\ \mathrm{s}$ or use the relation $y=-\frac{1}{2}gt^2+v_0t+y_0$ as follows
\[0=-\frac{1}{2}\left(9.8\right)t^2+20t+0\Rightarrow t\left(9.8t-40\right)=0\] 
\[\Rightarrow \left\{ \begin{array}{lcl}
t=0\ \ & , & \ \ initial\ time\ \  \\ 
t=\frac{40}{9.8}=4.08\ {\rm s} \ \ & , &\ \ final\ time \end{array}
\right.\] 
(e) Use the equation $v^2-v^2_0=-2g(y-y_0)$, between the throwing and landing points
\[v^2-v^2_0=-2g\left(y-y_0\right)\Rightarrow v^2-{20}^2=-2\times 9.8\times \left(0-0\right)\] 
\[\Rightarrow v=20\frac{\mathrm{m}}{\mathrm{s}}\] 
 

A boat travelling at $12\ \mathrm{miles/hr}$ goes due east for $30$ minutes and then turns $45{}^\circ $ toward the north and travels another $20$ minutes.
(a) What total distance does the boat travel?
(b) What is the magnitude of its displacement?
(c) What is the direction of its displacement?
(d) What is the magnitude of its average velocity?
(e) What is the direction of its average velocity?

(a) The boat has constant velocity so its acceleration is zero. Therefore, use the equation of the uniform motion as $x=vt$ where $x$ is the distance traveled. The sketch of the path is shown in the following

 

 


\[x_t=x_1+x_2=v_1t_1+v_2t_2=\left(12\frac{\mathrm{mile}}{\mathrm{hr}}\right)\left(\mathrm{30\ min+20\ min}\right)\left(1\frac{\mathrm{hr}}{\mathrm{60\ min}}\right)\] 
\[x_t=\ \left(12\frac{\mathrm{mile}}{\mathrm{hr}}\right)\left(\frac{1}{2}\mathrm{hr+}\frac{\mathrm{1}}{\mathrm{3}}\mathrm{hr}\right)\mathrm{=10\ mile}\] 
(b) Displacement is a vector that its initial position is at the first vector and its final position is at the final of the last vector. So as shown in the figure $x_t$ is the displacement vector. Use the Pythagorean theorem to find it as follows
$x^2_t={\left(x_1+x_2{\cos 45{}^\circ \ }\right)}^2+{\left(x_2{\sin 45{}^\circ \ }\right)}^2$
\[\Rightarrow x^2_t={\left(6+4{\cos 45{}^\circ \ }\right)}^2+{\left(4{\sin 45{}^\circ \ }\right)}^2\] 
\[x_t=\sqrt{77.94+8}\] 
\[\Rightarrow x_t=9.27\ \mathrm{mile}\] 
(c) The angle of the vector $\vec{R}=R_x\,\rm\hat i+R_y\,\rm\hat j$ with respect to the positive $x$ direction is determined by $\theta ={{\tan}^{-1} \left(\frac{\left|R_y\right|}{\left|R_x\right|}\right)\ }$. Therefore ,
\[{\tan \theta \ }=\frac{x_2{\sin 45{}^\circ \ }}{x_1+x_2{\cos 45{}^\circ \ }}=\frac{4{\sin 45{}^\circ \ }}{6+4{\cos 45{}^\circ \ }}\] 
\[\Rightarrow \theta =17.8{}^\circ \] 
So the resultant vector lies in the first quadrant of the coordinate system.
(d) Velocity is a vector quantity and defines as the displacement of the particle divided by the time interval that is $\vec{\overline{v}}=\Delta \vec{x}/\Delta t$so its magnitude is 
\[\left|\vec{\overline{v}}\right|=\frac{\left|{\vec{\mathrm{x}}}_{\mathrm{t}}\right|}{\Delta t}=\frac{9.27}{\left(\frac{1}{2}+\frac{1}{3}\right)}=11.1\frac{\mathrm{m}}{s}\] 
(e) As we see from the definition of the average velocity, it is a vector in the direction of the displacement vector so its angle relative to the positive $x$ axis is same as the displacement vector that is $\theta =17.8{}^\circ $.
 

A ball is thrown horizontally off the edge of a cliff with an initial speed of $15\ \mathrm{m/s}$ and hits the ground $3$ seconds later.
(a) How high is the cliff?
(b) How far does the ball travel horizontally before it hits the ground?
(c) What is the magnitude of the ball's displacement?
(d) What is the speed of the ball when it hits the ground?
(e) At what angle from the vertical is the ball moving when it hits the ground?

(a) In the kinematic problems, first consider an origin then use the kinematic equations to find the desired parameters. Let the point of throw as the origin. This is projectile motion. In projectile motions, the horizontal motion is under constant velocity and vertical is under uniform acceleration $a=-g$. So in the vertical stage use the kinematic equation $y=-\frac{1}{2}gt^2+v_{0y}t+y_0$ where $y_0$ is the initial height and solve for the landing point $-y$ ( since the ball hits the ground below the origin)
\[-y=-\frac{1}{2}\left(9.8\right){\left(3\right)}^2+0\left(3\right)+0\Rightarrow y=44.1\ \mathrm{m}\] 
(b) Use the following kinematic equation, we have $x=vt=15\times 3=45\ \mathrm{m}$
(c) Use the Pythagorean theorem and find the ball's displacement as shown in the figure.
\[{\left(\Delta r\right)}^2=x^2+y^2={45}^2+{\left(44.1\right)}^2=63\ \mathrm{m}\] 
(d) First calculate its components at the moment that the ball hits the ground then use the Pythagorean theorem to find its magnitude as follows
\[v=v_0+at\to \ \left\{ \begin{array}{c}
v_x=v_{0x}=15\frac{\mathrm{m}}{\mathrm{s}} \\ 
v_y=v_{0y}-gt=0-\left(9.8\right)\left(3\right)=-29.4\frac{\mathrm{m}}{\mathrm{s}} \end{array}
\right.\] 
\[\Rightarrow \left|v\right|=\sqrt{v^2_x+v^2_y}=\sqrt{{15}^2+{\left(-29.4\right)}^2}=33\frac{\mathrm{m}}{\mathrm{s}}\] 
(e) The angle with respect to the positive $x$ axis is determined by the following relation 
\[\theta ={{\tan}^{-1} \left(\frac{\left|v_y\right|}{\left|v_x\right|}\right)\ }={{\tan}^{-1} \left(\frac{15}{29.4}\right)\ }=27.03{}^\circ \] 
Since $v_y$ is a negative number and $v_x>0$, so the final velocity directed $27.03\,^\circ$ below the +x axis.


Kinematic
Category : Kinematic

MOST USEFUL FORMULA IN KINEMATIC:

Horizontal motion with constant acceleration:
\[v_x=v_0+a_xt\]
\[x=\frac 1 2 a_x t^2 +v_0t+x_0\]
\[v^2-v_0^2=2a\Delta x\]

Free falling motion:
\[v_y=v_0-gt\]
\[y=-\frac 1 2 gt^2+v_0t+y_0\]
\[v^2-v_0^2=-2g\Delta y\]


Number Of Questions : 11