Physics 1 Test 1 Solutions

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 1/28 PHYS 1441 (Fall 2023 White) Test-1 View Basic/Answers Note: The variables used in the below solutions are not the same as those used in your assignment. Begin Date: 9/14/2023 5:30:00 PM -- Due Date: 9/14/2023 6:50:00 PM End Date: 9/14/2023 6:50:00 PM Problem 1 - 1.2.9 : Suppose you take a deep breath, and your lungs fill with a volume of air equal to 1.75 L. Part (a) What is the mass of the breath, in grams, if the density of air is 1.29 × 10 -3 g/cm 3 ? The volume of an object of uniform density is where m is the mass in kg and is the density in kg/m . Therefore, where V is the volume in m . Plugging in numbers and converting units as needed, Problem 1 - 1.2.10 : A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder filled with water. Part (a) What is the density, in grams per cubic centimeter, of a rock that displaces of water? The volume of an object of uniform density is where is the mass in kilograms and is the density in kilograms per cubic meter. Therefore, where is the volume in cubic meters. Plugging in numbers and converting units as needed, Problem 1 - 1.2.11 : Suppose you have a coffee mug with a cylindrical shape (the radius is uniform, in other words). Part (a) What is its inside radius, in centimeters, if it holds 350 g of coffee when filled to a depth of 5.5 cm? Assume coffee has the same density as water, 1.00 g/cm 3 . V = m ρ m 3 ρ 3 m = Vρ 3 m = 1.75 ⋅ ⋅ 1.29 ⋅ 10 3 cm 3 10 −3 g/cm 3 m = 2.258 g 225 g 87.5 cm 3 V = m ρ m 3 m ρ ρ = m V V ρ = 225 g 87.5 cm 3 ρ = 2.571 g/cm 3

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 2/28 The volume of an object of uniform density is where m is the mass in kg and is the density in kg/m . The volume of a cylinder is where r is the radius of the cylinder in m and h is the height of the cylinder in m. Setting the volumes equal, Plugging in numbers and converting units as needed, Problem 2 - c3.1.1 : A particle moves from point A to point B. Let L be the straight-line distance between A and B, let R be the magnitude of the particle's displacement, and let S be the distance the particle traveled. Part (a) Which is always true? We learn from our textbook that the magnitude of the displacement from A to B equals the straight-line distance between A and B, no matter what the actual traveled distance is. Thus, by definition. The correct choice is, therefore, ... Part (b) Which is only sometimes true? The magnitude of the displacement from A to B, , equals the travel distance from A to B, , when and only when the particle's path is straight. Since the path can be curvy, the equality is not always true. The correct choice is, therefore, ... Part (c) Which can never be true? The magnitude of the displacement from A to B, , equals the travel distance from A to B, , when and only when the particle's path is straight. Since the path can be curvy, the equality is not always true. can be smaller than , but it can never be larger, since a straight path is the shortest path between two points. The correct choice is, therefore, ... Problem 3 - 3.1.34 : The position of a particle moving along the x axis is given by = ( 4.01 m) - ( 2.01 m/s) . Part (a) At what time (in seconds) does the particle cross the origin? The origin is where , so we need to solve for time in V = m ρ m 3 ρ 3 V = π h r 2 m 3 2 = π h m ρ r 2 r = m (ρπh) 0.5 r = R = 350 g 1 ⋅ 5.5 cm ⋅ π g/cm 3 0.5 R = 4.501 cm L = R L = R R S R = S R = S R S R = S R S R > S x(t) t x = 0

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 4/28 Calculating, Part (b) What is the average speed of the blast wave as a fraction of the speed of sound (343 m/s at sea level)? Dividing by vields Problem 3 - 3.1.37 : A woodchuck runs 15 m to the right in 4.01 s, then turns and runs 8.1 m to the left in 2.01 s. Part (a) What is the magnitude of the average velocity of the woodchuck in m/s? Recall that velocity is equal to displacement divided by time. Since displacement includes direction, we have to take into account that the woodchuck turns around and runs in the opposite direction by applying a negative sign to the second displacement. Note that whether we define to the right or to the left as the negative direction actually isn't terribly important in this case, as we are asked for a magnitude. This means that we will take the absolute value of whatever the displacements sum to anyway. However, as the magnitude is always positive, taking left as the negative direction is convenient since the displacement to the left is smaller; this gives us a positive result without needing to think about swapping the signs. The time, meanwhile, is always going to be positive so the two times will be added together. Putting this all together, the equation we find is: Part (b) What is its average speed in m/s? Here, we are asked for the average speed. Unlike velocity, speed does not care about direction. Therefore, we only need to take the total distance traveled and divide it by the total time it took to find the average speed. Problem 3 - 3.1.39 : A professional baseball pitcher throws a baseball at a speed of towards home plate, which is away. Part (a) At that speed, how long does it take, in seconds, for the ball to reach home plate? Convert the speed to meters per second. Relating the distance to the velocity and the time, or v = (23.5 km) 1000 m 1 km [(2) (60 s) + 15 s] v = 174.1 m/s = 343 m/s v sound = 0.5075 v v sound = v avg − d 1 d 2 + t 1 t 2 = 15 m−8.1 m 4.01 s+2.01 s = 1.146 m/s v avg = v avg + d 1 d 2 + t 1 t 2 = 15 m−8.1 m 4.01 s+2.01 s = 3.837 m/s v avg 130 km/h 18.4 m v = = 130 × × km h 1000 m 1 km 1 h 3600 s 36.11 m/s d = vt

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 5/28 Problem 3 - 3.1.40 : An airplane leaves Chicago and makes the 3000-km trip to Los Angeles in 4.5 hours. A second plane leaves Chicago 0.51 hours later and arrives in Los Angeles at the same time. Express your answers in units of km/h. Part (a) What is the speed of the first plane? Speed is equal to distance divided by time, so Calculating, Part (b) What is the speed of the second plane? The time of travel for the second plane is 0.51 h less than that for the first plane, so Calculating, Problem 4 - c3.3.5 : A rock is thrown straight upward. Neglect air resistance. Part (a) What is the direction of its acceleration as it is rising? The rock's acceleration is due to Earth's gravitational force, which is always pulling the rock downward, causing a constant downward acceleration. The rock's downward acceleration slows the rock down while it is ascending, reverses the rock's velocity at the trajectory's highest point, and speeds up the rock during its descent. But the acceleration due to gravity is always downward. The correct choice is, therefore, ... Downward Part (b) What is the direction of its acceleration at the moment it reaches its highest point? The rock's acceleration is due to Earth's gravitational force, which is always pulling the rock downward, causing a constant downward acceleration. The rock's downward acceleration slows the rock down while it is ascending, reverses the rock's velocity at the trajectory's highest point, and speeds up the rock during its descent. But the acceleration due to gravity is always downward. The correct choice is, therefore, ... Downward Part (c) What is the direction of its acceleration as it is falling? The rock's acceleration is due to Earth's gravitational force, which is always pulling the rock downward, causing a constant downward acceleration. The rock's downward acceleration slows the rock down while it is ascending, reverses the rock's velocity at the trajectory's highest point, and speeds up the rock during its descent. But the acceleration due to gravity is always downward. t = = d t 18.4 m 36.11 m/s t = 0.5095 s speed = 3000 km 4.5 hr speed = 666.7 km/h speed = 3000 km 4.5 hr − 0.51 hr speed = 751.9 km/h

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 7/28 • Note that the first portion of the acceleration graph is positive indicating that the same portion of the velocity curve should have a positive slope. • Note that the second portion of the acceleration graph is negative indicating that the same portion of the velocity curve should have a negative slope. • The area between the curve and above the horizontal axis in the first section is equal to the area between the curve and below the horizontal axis in the second section, so the increase in velocity should be followed by a decrease in velocity of equal magnitude. The correct choice is Problem 5 - c3.4.15 : A graph of acceleration versus time is shown. Part (a) Which of the following graphs best represents a velocity curve that is consistent with the acceleration curve in the above problem statement? The horizontal axes are scaled identically. • Recall that acceleration represents the slope of the velocity curve. • Note that the first portion of the acceleration graph is zero indicating that the slope of the same portion of the velocity curve should have a zero slope. In other words, that portion of the velocity curve should be a constant value. We don't know what that constant value is unless we are given the initial velocity. • Note that the second portion of the acceleration graph is negative indicating that the same portion of the velocity curve should have a negative slope. The correct choice is

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 8/28 Problem 5 - c3.4.16 : A graph of acceleration versus time is shown.

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 10/28 • The velocity curve indicates an initially non-zero velocity which has no impact on the acceleration curve. The correct choice is Problem 5 - c3.4.18 : A graph of velocity versus time is shown. Part (a) Which of the following graphs best represents an acceleration curve that is consistent with the above velocity curve? The horizontal axes are scaled identically. • Recall that acceleration represents the slope of the velocity curve. • Note that the first portion of the velocity graph has a constant positive slope indicating that the same portion of the acceleration curve should have a constant positive value. • Note that the second portion of the velocity graph is constant indicating that the same portion of the acceleration curve should be zero. • The velocity curve indicates an initially non-zero velocity which has no impact on the acceleration curve. The correct choice is

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 11/28 Problem 5 - c3.4.19 : A graph of velocity versus time is shown. Part (a) Which of the following graphs best represents an acceleration curve that is consistent with the above velocity curve? The horizontal axes are scaled identically. • Recall that acceleration represents the slope of the velocity curve. • Note that the first portion of the velocity graph is constant indicating that the same portion of the acceleration curve should be zero. • Note that the second portion of the velocity graph has a constant positive slope indicating that the same portion of the acceleration curve should have a constant positive value. • The velocity curve indicates an initially non-zero velocity which has no impact on the acceleration curve. The correct choice is

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 13/28 Part (c) Calculate the runner’s constant acceleration, in meters per second squared, after he crosses the finish line until he comes to rest. Because of the constant acceleration, a kinematic equation can be used. For this part of the problem, the final velocity, typically represented as in kinematics, corresponds to , which is equal to 0. The initial velocity, which we typically write as in kinematics, has the same value as from when the runner crossed the finish line. The variables involved in this part of the problem are as follows: In this case, we do not know the distance it takes for the runner to come to a stop, so we use the kinematic equation that does not include . We rearrange and solve for . It can be determined that and . These can be combined to determine an expression for in terms of variables in the problem. Plugging in: Problem 6 - 3.5.11 : A motorist in England is on a road that has a posted speed limit of and is traveling in excess of the speed limit by . A car pulls out of a parking place into the street in front of the motorist and stops there. The motorist immediately slams on the brakes and begins decelerating at a rate of . Part (a) By what conversion factor do you need to multiply a speed expressed in kilometers per hour in order to obtain the value of that speed expressed in meters per second? = v 1 a i t 1 v v 2 v 0 v 1 a v v 0 t = = = = a f = 0 v 2 = v 1 a i t 1 − = 5.5 s t 2 t 1 x a f v a a f = = = + at v 0 − v v 0 t −v 1 − t 2 t 1 = v 1 a i t 1 = t 1 2d a i v 1 v 1 = = = a i t 1 a i 2d a i 2da i √ a f = = = = −v 1 − t 2 t 1 − 2da i √ − t 2 t 1 − 2(100 m)(0.45 m/ ) s 2 √ 5.5 s −1.725 m/s 2 = −1.725 m/ a f s 2 = 50.2 km/hr v m = 10.3 km/hr v c d = 10.4 m a = 3.04 m/s 2

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 14/28 Let's begin by using dimensional analysis to convert hours to seconds. As for kilometers to meters, we know from our metric units that there are 1000 meters in a kilometer. From this, we can set up an equation to get our conversion factor. Therefore, our conversion factor is \[x = 0.2778 \frac{\text{m/s}}{\text{km/h}} \] Part (b) Select an expression for the minimum distance, , the motorist would need to come to a complete stop. To begin, let's look over our kinematic equations. We know that the initial velocity is the speed limit plus the amount the driver is speeding by. We also know the acceleration. We want to find the distance travelled. Finally, we know that the final velocity is zero since we want to find out how long until the car stops. We don't know or care about time right now. Given our known values and what we want to find, we should use the following kinematic equation. Now, we just need to plug in our known values to solve. The only thing be careful of is that since the acceleration is opposing the direction of motion to decelerate the car, our acceleration will be negative in this case. Part (c) In terms of the defined quantities, enter an expression for the time it would take the motorist to come to a stop after applying the brakes. To begin, we need to select the correct kinematic equation based on our known and unknown variables. We know that the initial velocity is given by the speed limit plus the amount the car is speeding by. We know the acceleration. We know that the final velocity is zero since we want to see how long until the car stops. We don't know the time, but we want to find out. Finally, we don't know or care about the distance travelled. Using this information, we can pick the best kinematic equation. Now we can use our known values to solve this equation for time. The only thing to be careful of is that deceleration is the same as backwards acceleration. Thus, we will have a negative acceleration. 1 h = ⋅ = 3600 s 60 min 1 h 60 s 1 min 1 km/h = m/s 1000 3600 1 km/h = 0.2778 m/s 1 = 0.2778 m/s km/h d c = + 2ad v 2 f v 2 i = + 2 (− ) 0 2 ( + ) v m v c 2 a b d c 0 = − 2 ( ) ( + ) v m v c 2 a b d c 2 = a b d c ( + ) v m v c 2 = d c ( + ) v m v c 2 2a b = + aΔt v f v i

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 16/28 To determine if the car hits the branch, we will need to find how far the car traveled before it stopped. To calculate this distance we will need to use the kinematic equation and apply it twice; once to the car while it is accelerating, and once to the car while it is stopping. For acceleration, we get the following equation: For deceleration, we get the following equation: To find the total distance traveled, we can sum these two equations. However, we have two problems with the deceleration equation in that we don't know either the final velocity or the final acceleration. Thankfully, we found a formula for the final velocity in part (a): To find the acceleration, we can use the same kinematic equation we employed to find the final velocity in part (a) and rearrange it to solve for the acceleration. Two key differences are that this time, there is an initial velocity (which is equal to the final velocity from when the car was accelerating) and that the final velocity is zero. This gives us the following formula: Now that we have formulas for both the deceleration and the final velocity, let's plug them into our equation for the distance traveled while the car is decelerating: Now that we have equations for both the distance traveled during the car's acceleration and its deceleration, we can add them together, simplify, and then plug in values to find the total distance the car travels. Since this distance is less than 550 meters, the car will stop before it hits the limb. The answer is therefore: No Part (c) How far, in meters, from the original location of the limb will the car be when it stops? In part (b) of this problem, we found that the car travels 375.0 meters before stopping. To find how far the car is from the limb when it stops, we need only subtract this from the distance the limb was from the car when it started accelerating. Problem 7 - 3.5.62 : Starting from rest, a cyclist travels in a straight line from point A to point B in 10.00 min. During the first 2.00 min of her trip, she maintains a uniform forward acceleration of 0.061 m/s 2 . She then travels at constant velocity for the next 5.00 min. Next, she slows with constant acceleration, so that she comes to a rest at point B 3.00 min later Part (a) What is the magnitude of her acceleration during the last 3.00 min in m/s 2 ? Let the positive direction be from A toward B. We'll start by finding the position and velocity after the first ( ). From and with and , d = t + a v 0 1 2 t 2 = d 1 1 2 a 1 t 2 1 = + d 2 v 1 t 2 1 2 a 2 t 2 2 = v 1 a 1 t 1 0 = + v 1 a 2 t 2 − = v 1 a 2 t 2 = −v 1 t 2 a 2 = −a 1 t 1 t 2 a 2 = d 2 + v 1 t 2 1 2 a 2 t 2 2 = ( ) + a 1 t 1 t 2 1 2 −a 1 t 1 t 2 t 2 2 = − a 1 t 1 t 2 1 2 a 1 t 1 t 2 = 1 2 a 1 t 1 t 2 = d tot + d 1 d 2 = + 1 2 a 1 t 2 1 1 2 a 1 t 1 t 2 = ( + ) 1 2 a 1 t 1 t 1 t 2 = 1.5 (20 s) (20 s + 5 s) 1 2 m/s 2 = 375.0 m d = 550 m − 375.0 m d = 175.0 m 2.00 min 120 s v = + at v 0 x = + t + a x 0 v 0 1 2 t 2 = 0 x 0 = 0 v 0

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 17/28 Applying the same equations to the middle ( ), with and , and , For the last ( ) the velocity becomes zero, so using with , , we have Calculating, Part (b) What is the distance from A to B in meters? From with , Calculating, Problem 7 - 3.5.65 : A police car waits in hiding slightly off the highway. A speeding car is spotted by the police car traveling at speed s = 25 m/s. At the instant the speeding car passes the police car, the police car accelerates forward from rest at a constant rate of a = 2.01 m/s 2 to catch the speeding car. Assume the speeding car maintains its speed. Part (a) Write an expression for the time it takes for the police car to catch the speeding car. Use the variables from the problem statement for your expression. In order to solve this problem, it's necessary to realize that because the two cars are involved in a chase, they will both travel the same distance and for the same amount of time. With that in mind, let's write an equation for the distance the speeding car travels using the kinematic equation . Next, let's write an equation for the distance the police car travels using the same kinematic equation. Since is the same in both equations, we can set the right side of each equation equal to one another. From there, we simply need to simplify to find the requested equation. Part (b) How long, in seconds, does it take for the police car to catch the speeding car? Since we found the equation we need to solve for the time in part (a), here we simply need to plug values into that equation and solve to get the answer. v = at = 0.061 (120 s) = 7.320 m/s m/s 2 x = a = 0.061 = 439.2 m 1 2 t 2 1 2 m/s 2 (120 s) 2 5.00 min 300 s = 439.2 m x 0 = 7.320 m/s v 0 a = 0 v = = 7.320 m/s v 0 x = + t = 439.2 m + (7.320 m/s) (300 s) = 2635 m x 0 v 0 3.00 min 180 s v = + at v 0 v = 0 = 7.320 m/s v 0 a = ∣ ∣ ∣ 0 − 7.320 m/s 180 s ∣ ∣ ∣ a = 0.04067 m/s 2 x = + t + a x 0 v 0 1 2 t 2 = 2635 m x 0 x = 2635 m + (0 + 7.320 m/s) (180 s) 1 2 x = 3294 m d = + vt + a d 0 1 2 t 2 d = t v s d = 1 2 a p t 2 d = d = t 1 2 a p t 2 v s = t 1 2 a p t 2 v s t = 2v s a p t = 2v s a p = 2(25 m/s) 2.01 m/s 2

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 19/28 Part (a) How many seconds after the second player reaches the ground does the first player reach the ground? Neglect air resistance. Take upward to be the positive direction, so and let the ground be the origin. At the highest point the velocity is zero, so with in , we have . Substituting this into with and results in Solving for time, is the time to reach the high point. Similarly, the time to fall from the high point can be calculated using with , , and , so that . Solving for time gives as the time to fall from the high point. Thus, the total time in the air is . Subtracting the shorter time from the longer time gives the difference in landing times: Calculating, Problem 8 - 3.6.38 : A hot-air balloon rises from ground level, moving straight upward at a constant speed of 5.01 m/s. At 20.1 s after liftoff, a sandbag is accidentally dropped from the balloon. Neglect air resistance. Part (a) How long, in seconds, does it take the sandbag to reach the ground from the moment it falls off the balloon? The height of the balloon at the moment the sadbag is released can be calculated using distance equals speed times time: Take upward to be the positive direction, so that the acceleration of the sanbag after release is . If the ground is the origin, then and in the equation , so Using the quadratic formula, Calculating the positive root, Part (b) How fast is the sandbag moving, in m/s, just before it hits the ground? Using , and recognising the speed as the magnitude of the displacement, we have Calculating, a = − 9.80 m/s 2 v = 0 v = + at v 0 = − at v 0 x = + t + a x 0 v 0 1 2 t 2 = 0 x 0 x = h h = −a + a = − a t 2 1 2 t 2 1 2 t 2 t = −2h/a √ x = + t + a x 0 v 0 1 2 t 2 = 0 v 0 = h x 0 x = 0 0 = h + a 1 2 t 2 t = −2h/a √ t = 2 −2h/a √ Δt = 2 − 2 −2 (0.81 m) −9.80 m/s 2 −2 (0.31 m) −9.80 m/s 2 Δt = 0.3101 s height = (5.01 m/s) (20.1 s) = 100.7 m a = − 9.80 m/s 2 x = 0 = 100.7 m x 0 x = + t + a x 0 v 0 1 2 t 2 0 = (100.7 m) + (5.01 m/s) t + −9.80 1 2 m/s 2 t 2 t = − (5.01 m/s) ± − 4 −9.80 (100.7 m) (5.01 m/s) 2 1 2 m/s 2 2 −9.80 1 2 m/s 2 t = 5.073 s v = + at v 0 v = 5.01 m/s + −9.80 (5.073 s) ∣ ∣ m/s 2 ∣ ∣ v = 44.71 m/s

12/12/23, 3:28 PM The Expert TA | Human-like Grading, Automated! https://usq45tx.theexpertta.com/Common/ViewAssignmentSolutionsV2.aspx 20/28 Problem 8 - 3.6.39 : An object is dropped from the roof of a building of height h . Neglect air resistance. Part (a) During the last 0.51 s of its fall, the object drops a distance h /3 before hitting the ground. What is h in meters? To begin, let us use the kinematic equation to write a formula for the entire fall. For the sake of simplicity, we will let down be the positive direction. Now let's repeat this process to get an equation for the first two-thirds of the fall. We know that the time it takes for the object to fall this far will be the total time minus the time for the final third of the fall . Right now, we have two unknowns ( and ) and want to find the height. To get the height by itself, we can take the square root of both the equations we have found so far and subtract them from one another. After we have the height by itself as the only unknown value, we can plug in values and solve to find the answer. Problem 9 - c2.2.6 : The diagram shows four vectors in the x - y plane. Part (a) Select all vectors with an x component of zero. Looking at the diagram, we can see that all of the vectors but #2 have some horizontal component in the x -direction. The correct answer is therefore: Vector 2 Part (b) Select all vectors with a y component of zero. x = + t + a x 0 v 0 1 2 t 2 h = g 1 2 t 2 t t f h = g 2 3 1 2 t − t f 2 t h − = t − t − h h 2 3 g 1 2 t f g 1 2 − = t − t + h 2 3 h g 1 2 g 1 2 t f g 1 2 1 − = h 2 3 t f g 1 2 = h t f g 1 2 1− 2 3 h = g t 2 f 1 2 1− 2 3 2 = 9.8 (0.51 s) 2 1 2 m/s 2 1− 2 3 2 h = 37.85 m