PES 1110 Exam I (Fall 2024) - Past Paper PDF

Here is the transcription of the provided text into a structured markdown format: ### PES 1110 Exam I, Fall 2024 - version 3 **Name:** Solutions **Score:** /33 = % Include your work. Answers without work receive no points. ### Circular Motion (2 points) 1. What is the relationship between th...

Here is the transcription of the provided text into a structured markdown format: ### PES 1110 Exam I, Fall 2024 - version 3 **Name:** Solutions **Score:** /33 = % Include your work. Answers without work receive no points. ### Circular Motion (2 points) 1. What is the relationship between the acceleration in circular motion, the speed of the object, and the radius of the circle? $\mathbf{a} = \frac{v^2}{R}$ 2. What is the direction of the acceleration vector? to center of circular path ### Kinematic Equations: Constant Acceleration (4 points) Give the time explicit expressions for velocity and position if acceleration is constant. $v(t) = \vec{a}t + \vec{v}_0$ $\vec{r}(t) = \frac{1}{2}\vec{a}t^2 + \vec{v}_0t + \vec{r}_0$ ### Calculus (2 points) 3. What is a Derivative (in words)? ratio of two very small quantities or slope of a plot 4. What is an Integral (in words)? Summation of many very small terms - (infinite number of infinitesimal things). or Area under curve of a plot ### Problem 1: Kinematic Plots (5 points) A runner is on a straight track. The graph below shows the position of the runner as a function of time. The graph shows position vs. time. The x axis measures time in seconds from 0 to 20. The y axis measures position in meters from -5 to 30. The graph shows position increasing linearly to 15 m at 5 s. Position is then constant until 12 s. The runner then moves back to 20 m at 15 s and stays there through 20 s. a) What is the average velocity of this plot? $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_0}{t_f - t_0} = \frac{20m - 20m}{20s - 0s} = \frac{0m}{20s} = 0$ $v_{avg} = 0$ b) What is the velocity at time t = 5 s? $v = slope = \frac{\Delta x}{\Delta t} = \frac{x_f - x_0}{t_f - t_0}$ $v= \frac{0m - 20m}{8s - 0s} = \frac{-20m}{8s} = -2.5 \frac{m}{s}$ $v=-2.5\frac{m}{s}$ c) What is the velocity at t = 15 s? $v = slope = \frac{x_f - x_0}{t_f - t_0}$ $v= \frac{20m - 0m}{16s - 12s} = \frac{20}{4} \frac{m}{s} = 5 \frac{m}{s}$ d) What is the runner's average acceleration? $a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_0}{t_f - t_0} = \frac{0 - (-2.5 \frac{m}{s})}{20s - 0s}$ $a = \frac{+2.5\frac{m}{s}}{20s} = 0.125 \frac{m}{s^2}$ e) Sketch an acceleration vs time plot for the car. It does not have to be exact. The graph shows an rough sketch of acceleration vs. time. The x axis measures time in seconds from 0 to 20. The y axis measures acceleration in m/s^2. There are two large spikes in acceleration around 7 s and 12 s, a large negative spike around 16 seconds, and 0 acceleration elsewhere. ### Problem 2: 1-D Motion (5 points) A car starts at the beginning of a straight track with an initial velocity. The car is accelerating at a constant rate of $a=10 \frac{m}{s^2}$. 20 meters away, at the end of the track, is a ravine. A ramp is also at the end of the track so the car can jump the ravine. In order to clear the ravine, the car needs a minimum speed of $40 \frac{m}{s}$ at the ramp. a) Find minimum initial velocity of the car to clear the ravine ($v_0$ = ???) $v_f = at_f + v_0$ $D = \frac{1}{2}at_f^2 + v_0t_f$ find $t_f = \frac{v_f-v_0}{a}$, plug into $D$ $aD = \frac{1}{2}a\frac{(v_f-v_0)^2}{a^2} + v_0\frac{(v_f - v_0)a}{a}$ $aD = \frac{1}{2}\frac{(v_f^2-2v_0v_f+v_0^2)}{a} + \frac{v_0v_f - v_0^2}{a}$ $aD = \frac{1}{2}v_f^2 - \frac{1}{2}v_0^2$ $aD = \frac{1}{2}v_f^2 - \frac{1}{2}v_0^2$ (Over for part b and more work space) $-2aD = -v_f^2 + v_0^2$ $v_0^2 = +v_f^2 - 2aD$ $v_0 = \sqrt{v_f^2 - 2aD}$ $v_0 = \sqrt{(40 \frac{m}{s})^2 - 2(10 \frac{m}{s^2})(20m)}$ $v_0 = 34.6 \frac{m}{s}$ b) A third car (Car #3) has a velocity that is given as a function of time. Find Car #3's acceleration as a function of time (1 point)? If $v_3 = \alpha t^5$ then $a_3 = ?$ $(\alpha = 0.2 \frac{m}{s^6})$ $a_3 = \frac{dv_3}{dt} = \frac{d \alpha t^5}{dt}$ $a_3 = 5\alpha t^4$ ### Circular Motion (5 points) On an in icy road, the road makes a turn in the shape of a quarter circle. The quarter circle has a radius of R = 30 meters. If a car accelerates faster than $a = 14 \frac{km}{hour \cdot second}$ they slide. What is the shortest time a car could make this turn in? (give answer in seconds) $v = \frac{distance}{time} = \frac{\frac{\pi}{2}R}{t_f} \implies t_f = \frac{\frac{\pi}{2}R}{v}$ find $v$... $a = \frac{v^2}{R} = 14 \frac{km}{hr \cdot s}$ (max possible acc, ... max possible $v$) $v = \sqrt{aR} \implies t_f = \frac{\frac{\pi}{2}R}{\sqrt{aR}} = \frac{\frac{\pi}{2}R}{\sqrt{aR}\frac{1}{2}}$ (use other side of for more work) $t_f = \frac{\frac{\pi}{2} \sqrt{R}}{\sqrt{a}}$ * $[R] = [m]$ if $t_f$ is in sec. $ [\alpha] = \frac{km}{}$ need to convert $a$ $a = 14 \frac{km}{hrs} \cdot \frac{1000m}{1 km} \cdot \frac{1 hr}{60 min} \cdot \frac{1 min}{60 s}$ $a = 3.89 \frac{m}{s^2}$ $t_f = \frac{\frac{\pi}{2} \sqrt{R}}{\sqrt{a}} = \frac{\frac{\pi}{2} \sqrt{30\ m}}{\sqrt{3.89 \frac{m}{s^2}}}$ $t_f = 4.36 s$ ### Problem 4: Relative Velocity and Vector Notation (5 points) On a windy day Bob is standing on the ground. He is watching an airplane, a balloon, and a car. He uses a Cartesian coordinate system where the North direction is the y-direction and the East direction is the x-direction (up is the z-direction, but nothing is moving up). Bob measures the plane's velocity to be: $\vec{v}_{p/B} = 80 \frac{m}{s}\hat{i} - 90 \frac{m}{s}\hat{j}$ Bob measures the balloon's velocity to be: $\vec{v}_{b/B} = 10 \frac{m}{s}\hat{i}$ Bob measures the car's velocity to be: $\vec{v}_{c/B} = 20 \frac{m}{s}\hat{i} - 20 \frac{m}{s}\hat{j}$ a) Draw a diagram depicting this problem. Visually represent all velocity vectors with arrows (scale the relative magnitudes of the vectors). Be sure to indicate a coordinate system. (1 point) There is a diagram depicting vectors on a coordinate systems. The vectors are labeled Vp/B, Vb/B, and Vc/B. b) What is the plane's velocity relative to the car? (2 points) (What is the velocity the car sees the plane moving at? Use vector notation.) $\vec{v}_{p/c} = \vec{v}_{p/B} - \vec{v}_{c/B}$ $= (80 \frac{m}{s}\hat{i} - 90 \frac{m}{s}\hat{j}) - (20 \frac{m}{s}\hat{i} - 20\frac{m}{s}\hat{j})$ $= (80\frac{m}{s} - 20\frac{m}{s})\hat{i} + (-90 \frac{m}{s} + 20\frac{m}{s})\hat{j}$ $\vec{v}_{p/c} = 60 \frac{m}{s}\hat{i} - 70 \frac{m}{s}\hat{j}$ c) According to the car, what would the angle the plane's velocity would make measured from the x-axis? Draw a diagram of the vector you are considering and label the angle you are finding. (2 points) $tan\theta = \frac{70 \frac{m}{s}}{60 \frac{m}{s}} = \frac{7}{6}$ $\theta = arctan(\frac{7}{6})$ $\theta = 49.4^{\circ}$ ### Problem 5: Projectile Motion (5 points) A cannon is a distance of D = 50 meters away from a wall. The wall has a height of 10 meters. The cannon ball is fired from the ground and makes an angle $\theta = 10^{\circ}$ with the ground. a) Draw a diagram depicting the situation. Label all given quantities in the diagram. There is a diagram showing a cannon angled at 0 degrees firing a cannonball at a point on a wall 10 m high and 50 m away. a) What is the initial speed the cannon ball needs to hit the wall STRAIGHT ON (angle of the velocity of the ball measured relative to the ground is zero when it hits the wall)? $v_x = v_0cos\theta$ $v_y = -gt +v_0sin\theta$ $x = v_0cos\theta t$ $y = -\frac{1}{2}gt^2 + v_0sin\theta t$ at $t_f$, $v_x = v_0cos\theta$, but $v_y=0$ (only in x-dir.) $v_{yf} = 0 = -gt_f + v_0sin\theta$ $\therefore t_f = \frac{v_0sin\theta}{g}$ we know that at $t_f$, $x_f=D$ $D = v_0cos\theta t_f = v_0cos\theta (\frac{v_0sin\theta}{g})$ $D = \frac{v_0^2}{g} cos\theta sin\theta$ (Solve for $v_0$) $v_0 = \sqrt{\frac{gD}{cos\theta sin\theta}} = \sqrt{\frac{(9.81\frac{m}{s^2})(50m)}{cos(10^{\circ})sin(10^{\circ})}}$ $v_0= 53.6 \frac{m}{s}$