Velocity Calculations 1. A traveler drives 568 km in 7.2 h. What is the average speed for the trip? 79 km/h 2. If you run with an average speed of 12.0 km/h, how far will you go in 3.2 min? 0.64 km 3. If the average speed of your private jet is 8.0 x 102 km/h, how long will it take you to travel 1.8 x 103 km? 2.3 h 4. The following distances and times for consecutive parts of a trip, made by a red ant on the prowl for food, were recorded by different observers. There is considerable variation in the precision of their measurements. A. 4.56 m in 12 s B. 3.4 m in 6.89 s C. 12.8 m in 36.235 s a) What total distance did the ant travel? 20.8 m b) What was the total time for the trip? 55 s c) What was the average speed of the ant? 0.38 m/s or 38 cm/s 5. Light travels with a speed of 3.00 x 105 km/s. How long will it take light from a laser to travel to the moon (where it is reflected by a mirror) and back to Earth? The moon is 3.84 x 105 km away from the Earth. 2.56 s Average Velocity Calculations 1. A high school bus travels 240 km in 6.0 h. What is its average speed for the trip? (in km/h) vav = d ÷ t vav = 240 ÷ 6 vav = 40. km/h 2. A black widow spider chasing its mate (which it will eat) travels across a driveway 3.6 m wide with a speed of 12 cm/s. How long will it take to cross the driveway? vav = d ÷ t t = d ÷ vav t = 3.6 ÷ 0.12 t = 30. s 3. Joe Blow, one of the many stars of the local basketball team, steals the ball and runs the length of the court in 8.5 sec at a speed of 5.0 m/s. How long is the court? d = vt where v = 5 m/s, t = 8.5 s d = 5 x 8.5 d = 43 m 4. The average speed of a runner in a 400.0 metre race is 8.0 m/s. How long did it take the runner to complete the race? vav = d ÷ t t = d ÷ vav t = 400 ÷ 8 t = 50. s 5. If a car is traveling at 25 m/s, how far does it travel in 1.0 hour? d = vt d = 25 m/s x 3600 s d = 9.0 x 104 m 6. A caterpillar travels across the length of a 2.00 m porch in 6.5 minutes. What is the average velocity of the caterpillar in m/s? vav = d ÷ t v= 2.00 ÷ 390 s v = 5.1 x 10-3 m/s 7. A motorist traveling on a straight stretch of open highway sets his cruise control at 90 km/h. How far will he travel in 15 minutes? d = vt d = 90 km/h x 0.25 h d = 23 km 8. A motorcycle travels 90.0 km/h. How many seconds will it take the motorcycle to cover 2.00 x 103 m? t = d ÷ vav t = 2.00 x 103 m ÷ 25 m/s t =80 s 9. A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 280.5 m from the wall and the sound of her voice travels at 340 m/s at that location. How long after she shouts will she hear her echo. t = d ÷ vav t = 561 m ÷ 340 m/s t = 1.65 s 10. A woman from Pasadena makes a trip to a nearby shopping mall that is located 40 km from her home. On the trip to the mall she averages 80 km/h but gets a speeding ticket upon her arrival. On the return trip she averages just 40 km/h. What was her average speed for the entire trip ? v = 53.3 km/h (determine total distance traveled and divide by total time) 11. A cross-country racecar driver sets out on a 100 km race. At the halfway marker (50 km), her pit crew radios that she has averages only 80 km/h. How fast must she drive over the remaining distance in order to average 100 km/h for the entire race? v = 133 km/h 12. A supersonic jet travels once around the earth at an average speed of 1.6 x 103 km/h. Its orbital radius is 6.5 x 103 km. How many hours does the trip take? d = vt where d = circumference of path (2πr), v = 1.6 x 103 km/h d÷v = t time will be in hours if we use d in km, and v in km/h 2πr÷v = t t = 2 x π x 6.5 x 103 ÷ 1.6 x 103 t = 25.5 h = 26 h Worksheet - Distance-Time Graphs Examine this graph carefully to answer questions 1 and 2. 1. How far is the truck from its starting point after: (a) 10 s 300 m s 300 m (c) 30 s 200 m s 650 m (e) 50 s 800 m (b) 15 (d) 43 2. What is the truck’s velocity in each of the intervals A through E? A 30 m/s B C -10 m/s D 40 m/s E 20 m/s 0 m/s 3. Beside the graph, describe briefly the kind of motion that is taking place in each of the situations represented by the following displacement vs. time graphs. a) object travels fast then changes to slower speed b) object travels fast then stops c) object travels slowly then changes to higher speed, stops then speeds off quickly 4. Beside the graph, describe briefly the motions represented by each of these graphs. If the speed is changing, state whether it is increasing or decreasing. a) object travels faster and faster (accelerating) b) object travels fast then slows down (decelerating) c) object travels slowly then changes to higher speed (accelerates), then slows down (decelerates) 5. Beside the graph, describe the motion represented by each of the following displacement vs. time graphs. a) object travels faster, stops then travel fast back to starting point b) object travels fast then immediately travels back part way, then forward and back several times c) object travels slowly then changes to higher speed (accelerates), then slows down (decelerates) to a stop and reverses direction coming back to the starting point Use the following graph for question 6. 6. a) From the graph above, calculate the average speed for the entire 22 s. v=d/t or v = d ÷ t v = 25 ÷ 22 v = 1.14 m/s b) Find the average speed for each of the following time intervals: (i) 0 s to 2 s v=d÷t v=5÷2 v = 2.5 m/s (ii) 4 s to 8 s v=d÷t v=0÷4 v = 0 m/s (iii) 14 s to 18 s v=d÷t v = (19-13) ÷ 4 v = 1.5 m/s (iv) 6 s to 17 s v=d÷t v = (15-5) ÷ 11 v = 0.91 m/s Constant Acceleration Problems = FUN! 1. A skier accelerates at 1.20 m/s2 down an icy slope, starting from rest. How far does he get in a) 5.0 s? 15 m b) 10.0 s? 60. m c) 15.0 s? 135 m 2. What is the acceleration of an object that accelerates steadily from rest, traveling 10.0 m in 10.0 s? d = vit + 1/2at2 d = 1/2at2 since vi = 0 m/s a = 2d/t2 a = 2(10.0 m)/(10.0 s)2 a = 0.20 m/s2 3. How long does it take an airplane, accelerating from rest at 5.00 m/s2, to travel 360 m? 12 m/s 4. A body moving with uniform acceleration has a velocity of 11.5 cm/s when its x coordinate is 5.0 cm. If its x coordinate 6.0 s later if -5.0 cm, what is its acceleration? -0.044 m/s2 use d=vit + 1/2at2 with vi = +0.115 m/s, d=-0.10 m 5. The initial velocity of a body is 4.60 m/s. What is its velocity after 5.50 s if it: a) accelerates uniformly at 3.00 m/s2 21.1 m/s b) accelerates uniformly at -3.00 m/s2? (that is it accelerates in the negative x direction) –11.9 m/s 6. A jet plane lands with a velocity of 115 m/s and can accelerate at a maximum rate of 5.25 m/s2 as it comes to rest. a) From the instant it touches the runway, what is the minimum time needed before it comes to rest? t = 21.9 s b) Can this plane land on a small tropical runway airport where the runway is 0.950 km long? d = vit + 1/2at2 d = (115 m/s)(21.9 s) + 1/2(-5.25)(21.9 s)2 d = 1260 m therefore, no, the plane cannot land on this runway 7. A body is traveling at 18.0 m/s and comes to rest after undergoing a uniform negative acceleration for 40.0 m. a) What is the acceleration of this body? –4.05 m/s2 b) How long does it take to come to rest? 4.44 s 8. A car enters a tunnel at 24 m/s and accelerates steadily at 2.0 m/s2. At what speed does it leave the tunnel, 8.0 s later? 40. m/s 9. A runner accelerates uniformly form rest at 1.40 m/s2 for 8.00 s. a) What is her final velocity? 11.2 m/s b) What is her average velocity? 5.6 m/s c) How far does she travel? 44.8 m 10. A car accelerates from rest at 6.65 m/s2. How far does it get between 10.0 and 14.0 s. 319.2 m Graphing in Kinematics 2 - Practice Examine this graph carefully to answer questions 1. 1. Above is a graph representing the motion of an object. For each section of that motion, labeled A through E, calculate the acceleration of the object. A 0 m/s2 B -2.0 m/s2 C 0 m/s2 D 2.5 m/s2 E -2.0 m/s2 2. How long will it take a truck moving at 35 m/s to stop if it accelerates at -5.0 m/s2? (show work) 7.0 s 3. A plane upon landing accelerates at -1.5 m/s2 for 1.0 minute until it stops. How fast was it going before it started to slow down? (show work) 90. m/s 4. A car hitting a tree loses 40.0 m/s in 0.100 s. What is its acceleration? (show work) -400. m/s2 5. Classify these graphs as representing constant, increasing, or decreasing acceleration. (a) (b) increasing acceleration acceleration decreasing (c) (d) constant acceleration constant acceleration of zero 6. This graph represents the motion of a truck that begins to move from rest. From the graph, calculate the average acceleration of the truck over the following time intervals. (show work) (a) 0 s to 150 s. (b) 150 s to 225 s. (c) 150 s to 350 s. (d) 0 minutes to 6 minutes. More Graphing Practice - Constant Acceleration Examine the v vs t graph above. Based on the motion represented by the graph, fill in the chart below: Time (s) Velocity (m/s) Acceleration (m/s2) Displacement (m) 0.0 4 - - 1.0 4 0 4 2.0 2 -2 7 3.0 0 -2 8 4.0 -2 -2 7 5.0 -2 0 5 6.0 0.5 2 4 7.0 3 2 6 8.0 1 -2 8.5 9.0 -1 -2 8 10.0 -3 -2 6 Examine the v vs t graph above. Based on the motion represented by the graph, construct the corresponding a vs t graph and d vs t graph on your own graph paper: Now refer back to the graphs and then 1. Draw a series of a diagrams of a car, person, ...or whatever moving with the motion depicted in the graph. Label the starting position in the diagram and be sure to point out and label what is happening to the object’s motion. 2. Along with your picture, right a series of statements specifically describing the characteristics of the object’s motion. (Is it speeding up or slowing down? Moving forward or backward? Accelerating or “decelerating”?) Car is going constant speed (no acceleration between 0 and 1 second), then slows down (decelerating from 1 second to 3 seconds). It stops at 3 seconds (see zero velocity), then comes back towards its starting point until it reaches 2 m/s (decelerating due to the direction from 3 to 4 seconds). Between 4 and 5 seconds it travels back towards its starting point at 2 m/s. The car then accelerates from 5 to 7 seconds, first reducing its speed towards starting point and then heading in the opposite direction increasing its speed. It hits 3 m/s at 7 second mark. From there the car slows down to zero m/s between 7 and 8.5 seconds. It then heads back towards its starting point from 8.5 to 10 seconds at a faster rate (but decelerating since it is coming back to its beginning as direction is negative) until it reaches 3 m/s towards the start point. Even More Graphing Practice - Constant Acceleration Complete the work on a separate piece of paper and attach it to this sheet... Examine the v vs t graph above. Based on the motion represented by the graph, answer the following questions (SHOW YOUR WORK)... 1. What is the acceleration of the object? Slope is –2.0 m/s2 2. What is the object’s displacement after 5 seconds? Area above x axis is 50. m 3. What is the object’s displacement after 10 s? Area above x axis is 56.25 m, area below is 6.25 m so therefore displacement is 50. m distance traveled after 10 seconds? Add the areas = 63.5 m 4. What is the equation for the line? v = -2.0t +15 Examine the v vs t graph above. Based on the motion represented by the graph, answer the following questions (SHOW YOUR WORK)... 5. What is the acceleration of the object between t = 0 and t = 8 s? Slope is 18.75 m/s2 ...between t = 8 s and t = 12 s? Slope is 0 m/s2 ...between t = 12 s and t = 20 s? Slope is –12.5 m/s2 6. What is the displacement of the object after 8 s? Area is 200 m approx...after 12 s? Area is 600 m ...after 20 s? Area is 1000 m note that v = 0 m/s at 2.67 m/s; these values are close approx. 7. What is the distance traveled by the object after 8 s? Add areas; Area is 333 m...after 12 s? Area is 733 m...after 20 s? Area is 1133 m 8. What is the equation for the line segment A? v = 18.75t - 50...line segment B? v = 100...line segment C? v = -12.5t +250 Acceleration Problems 1. A skier accelerates at 1.20 m/s2 down an icy slope, starting from rest. How far does she get in 5.0 s? 15 m 2. What is the acceleration of an object that accelerates steadily from rest, traveling 10 m in 10 s? 0.20 m/s2 3. A car enters a tunnel at 24 m/s and accelerates steadily at 2.0 m/s2. At what velocity does it leave the tunnel, 8.0 s later? 40 m/s 4. How long does it take an airplane, accelerating from rest at 5.0 m/s2, to travel 360 m? 12 s 5. The driver of a truck moving at 18 m/s throws on its brakes and stops it in 4.0 s. What is the acceleration of the truck? -4.5 m/s2 6. A motorcycle stuntman accelerates from rest to a maximum velocity of 35.2 m/s at the top of the take-off ramp, then swoops up and over 20 cars. Calculate how long it takes him to accelerate, at an acceleration of 8.8 m/s2. 4.0 s 7. Here is the velocity-time graph of a trip on a bicycle. How fast is the bicycle moving at each of the following time? a) 4 s 4 m/s b) 6 s 6 m/s c) 10 s 3.4 m/s d) 12 s 2 m/s What is the acceleration of the bicycle at each of these times? a) 2 s 0 m/s2 b) 5 s 1.0 m/s2 c) 7 s -0.67 m/s2 d) 14 2 s 0.75 m/s 8. If an electron inside a TV tube accelerates from rest to 1/10c (where c is the speed of light), in a space of 5.0 cm, what is its acceleration? c = 3.00 x 108 m/s therefore 1/10c is 3.00 x 107 m/s ; a 15 2 = 9.0 x 10 m/s 9. Two runners accelerate uniformly from rest at 1.40 m/s2 for 8.00 s. a) What is their final velocity? 11.2 m/s b) What is their average velocity? 5.6 m/s c) How far do they run? 44.8 m 10. A car accelerates from rest at 6.00 m/s2. How far does it get between 10.0 s and 15.0 s? At 10 s d = 300 m, at 15 s d = 675 m, therefore d between 10 and 15 seconds is 375 m 11. A skier accelerates steadily down a hill from 3.50 m/s to 11.40 m/s in 4.20 s. a = 1.88 m/s2 a) What is the average velocity for the trip? 7.45 m/s b) What is the displacement? 31.3 m 12. Jack and Jill ran down the hill. Both started from rest and accelerated steadily. Jack accelerated at 0.25 m/s2 and Jill at 0.30 m/s2. After running for 20.0 s, Jill fell down. a) How far did Jill get before she fell? 60 m b) How far had Jack traveled when Jill fell? 50 m c) How fast was Jack running when Jill fell? 5 m/s d) How long was it after Jill fell that Jack ran into her and broke his crown (to the nearest second)? 1.9 s longer 13. The graph shows the motion of a small bird flying down a long, straight, narrow cage. a) What is the bird’s acceleration for the first 10 s of the trip? Nearly a straight line, so 0.5 m/s2 b) At what time is the bird’s acceleration at its maximum? Find steepest part of graph, at about 20-23 s c) What is this maximum acceleration? About 1.3 m/s2 d) At what time(s) is the bird’s acceleration zero? At the top of the curve, so about 29 s e) At what time(s) is the bird’s velocity zero? At time zero. 14. Spiderman is crawling up a building at the rate of 0.50 m/s. Seeing Spiderwoman 56 m ahead of him; he accelerates at the rate of 2.3 m/s2. a) How fast will he be moving when he reaches Spiderwoman? 16.1 m/s b) How much time will it take to reach Spiderwoman? 6.8 s c) When he reaches Spiderwoman, Spiderman discovers she is a Black Widow and, as you know, Black Widows eat their mates! He is 200.0 m from the road below. How long will it take to fall to the safety of the road, if he drops with an acceleration of g = 9.80 m/s2? 6.4 s d) Why will Spiderman not be killed by the fall? He has spider-like senses and can whip out that web from his wrist to slow himself down Acceleration Due To Gravity Remember to use the value of a = 9.80 m/s2 for these problems. 1. A rock is dropped from the 100th floor of a tall building (350 m above the ground). How long does it take to fall to the ground below and with what speed will it hit the ground at? 8.45 s; 83 m/s 2. A ball is thrown straight upwards and reaches a maximum height of 95 m. With what initial velocity was it thrown? 43.2 m/s 3. A baseball is throw straight upwards at 28.0 m/s. a) To what maximum height will it reach? 40 m b) How long will it remain in the air (from being thrown upwards to being caught)? 5.7 s 4. A mischievous person is on an air balloon at a height of 750 m above the ground. He sees his archenemy standing below the balloon and throws a 10.0 kg anvil straight downward with an initial velocity of 15 m/s. With what speed does it hit the unaware enemy? 122 m/s How long was it before the archenemy was hit? 10.9 s 5. A pebble is thrown vertically downward from a bridge with an initial velocity of 10.0 m/s and strikes the water in 2.0 s. a) Find the velocity with which the pebble strikes 30. m/s and b) the height of the bridge. 40. m 6. A person throws a ball upward with an initial velocity of 15.0 m/s. Calculate: a) how high it goes. 11.5 m b) how long the ball is in the air before it comes back to his hand. 3.06 s 7. A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth? 1.5 s 8. A hammer falls out of a helicopter from a height of 3000.0 m. What velocity will the hammer reach just before it hits the ground if it continues to accelerate downwards? 242 m/s How long will it take to hit the ground? 24.7 s 9. A boy shoots a rock straight up with an initial velocity of 16 m/s. He quickly reloads and shoots another rock in the same way 2.0 s later. a) At what time and at what height doe the rocks meet? Pretty tricky, eh? So, the first rock takes 1.63 seconds to reach max height of 13.1 m. That means it has already fallen 0.37 seconds before the second rock is fired. The other rock is following the same pathway (and has same data). Consider the height and velocity of the first rock when the second rock is fired i.e. after it has fallen for 0.37 seconds. The first rock would be already traveling downward with a velocity of 3.63 m/s and has fallen 0.67 m (now at a height of 12.4 m off of the ground). Here's the trick. Given the new set up, the time for the rocks to meet would have to be the same AND the displacements would be related by the difference in height! So you have to get two equations and solve them as a system - a little mathematics from Math 10! Equations are (using d = vit + 1/2at2): Rock 1 12.4 - x = 3.63t + 1/2(9.80)t2 watch the signs!!!! Rock 2 x = 16t + 1/2(-9.80)t2 the x represents the distance that rock 2 will go up and the 12.4 – x represents the distance that rock 1 will fall before they meet. Put these together and reduce to find the value of t!! t = 0.63 s from when Rock 2 is fired and they meet at a height of 8.1 m b) What is the velocity of each rock when they meet? Work from the time Rock 1 9.8 m/s Rock 2 9.8 m/s 10. A water balloon is dropped from the top of a tower, 200.0 m off the ground. An alert archer at the base of the tower sees the balloon and shoots an arrow straight up toward the balloon 5.0 s after the balloon is dropped. The arrow's initial velocity is 40.0 m/s. Where does the arrow intercept the balloon? Draw a little picture of this. Think about the water balloon for a moment. Its initial velocity is 0 m/s but after 5.0 seconds it will have fallen 122.5 m (so it is now at a height of 200 - 122.5 = 77.5 m with a velocity of 49 m/s downwards). Then the arrow is fired. The arrow will go upwards but will travel some distance less than 77.5 m (because the balloon is still falling). But the time for the two objects to reach each other will be the same. Again you need two equations. One for the balloon dropping and one for the arrow traveling upwards. Plus you need to consider the effect of that 5.0 second time difference: Equations are (using d = vit + 1/2at2): Balloon Arrow 77.5 - x = 49 t + 1/2(9.80)t2 x = 40t + 1/2(-9.80)t2 watch the signs!!!! the x represents the distance that arrow will go up and the 77.5 - x represents the distance that water balloon will fall before they meet. Put these together and reduce to find the value of t!! t = 0.87 s (the arrow only reaches a height of 31.1 m before it encounters the balloon) Are there other ways to do Questions 9 and 10? Probably, I just did it my way! Challenge Problem 1. Sam challenges Jean to a race. Sam can run at 4.0 m/s. Jean can run at only 3.0 m/s. So Sam gives Jean a 20.0 m head start. a) How long does it take for Sam to catch Jean? 20 s b) How far from Jean’s starting point will they have traveled by the time Sam catches Jean? 60 m c) How far ahead of Jean will Sam be after 35 seconds into the race? 15 m Sam d + 20 = 4.0 t Jean d = 3.0 t Challenge Problem #2: 2. Three brothers decide to have a 200.0 m race. To be fair the youngest brother is given a 50.0 metre head start. The other two brothers run the full 200.0 m but the eldest gives the middle brother a 10.0 second head start. The youngest can run at 2.5 m/s, the middle brother at 4.0 m/s, and the eldest at 4.5 m/s. a) At what time does the eldest brother catch the youngest brother? 25 s b) How far ahead, or behind, is the middle brother at this time? Middle is at 140 m. Youngest and eldest are both at 112.5 m. This means that the middle brother is 27.5 m ahead (since he was given the 10 second head start). c) Who wins the race, and by how much? The middle brother by 4.4 seconds d) Who is the last, and by how much? Youngest was last by 15.6 seconds t Eldest d + 50 = 4.5 t Middle d = 4.0 t Youngest d = 2.5 Youngest takes 70 s to finish (this is 60 seconds to run plus the 10 second time delay) Middle takes 50 s to finish (remember he was the only runner for 10 seconds!) Eldest takes 54.4 s to finish (44.4 seconds of running plus the 10 second time delay!) Extra challenge: Their father starts at the other end of the 200 m run, intending to run in the opposite direction, 5.0 seconds after the eldest son. He can run at 6.0 m/s. At what time does he pass the last of his sons? 21.9 s after the eldest starts to run Eldest d = 4.5(t+5) Dad 200-d = 6.0t I am assuming that the eldest is the last of the sons since the distance that they meet is the 98.6 m mark and the other two brothers were already farther (youngest is at 105 m while middle is at 128 m at the 21.9 s mark)

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