#1) When the acceleration is constant, we can easily find the velocity acquired

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#1) When the acceleration is constant, we can easily find the velocity acquired as a function of time. Velocity acquired = acceleration X time What is the speed of a motorcycle after 10 seconds if the acceleration is 2 m/s^2. Question 1 options: 2 m/s 10 m/s 20 m/s 40 m/s #2) When the acceleration is constant, we can easily find the the final velocity of an object. Total Velocity = Initial velocity + acceleration X time You are traveling at 2 m/s on the 405 freeway. You are behind a car that is traveling at 3 m/s. Thus you accelerate at .2 m^2 for 5 sec. Will you be able to pass the car and how much faster or slower than them are you going? Question 2 options: No. You will not pass him since your final speed is 2 m/s. No since your final speed is 3 m/s and that is the same speed as the driver. Yes since you have accelated and your speed is 3 m/s. Yes since you have accelerated and your speed is over 3 m/s. #3) As an example, think about your automobile. You step on the gas and accelerate uniformly. At the end of 5 seconds, you are moving at 25 m/s. You then step on the brakes and end up stopping in 5 seconds. Your distance as a function of time is given in the table. What is the average velocity between 2 and 3 seconds? Time(s).|.Location(m).|.Velocity(m/s).|.Acceleration (m/s/s) 0_________0 m_____ 1_________5 m_____ 2________15 m_____ 3________30 m_____ 4________50 m_____ 5________75 m_____ 6_______100 m_____ 7_______120 m_____ 8_______135 m_____ 9_______145 m_____ 10______150 m______ 11______150 m______ To work this problem, fill out the table given above with the time, distance and speed. Use the definition of average velocity given earlier in this exercise and recognize that speed and velocity are the same for one dimensional motion. Average Velocity = ( location 2 - location 1 ) / total time = ( x2 - x1 ) / ( t2 - t1 ) Then answer the question below to check your answer. The next problem will give you the complete table so that you can check your answer. Question 3 options: 10 m/s 15 m/s 20 m/s 25 m/s #4) Now let us look at the acceleration from the table. What is the average acceleration between 2 and 3 seconds? Time(s).|.Location(m).|.Velocity(m/s).|.Acceleration (m/s/s) 0_________0 m______0 m/s__________ 1_________5 m______5 m/s__________ 3________30 m_____15 m/s__________ 4________50 m_____20 m/s__________ 5________75 m_____25 m/s__________ 6_______100 m_____25 m/s__________ 7_______120 m_____20 m/s_________ 8_______135 m_____15 m/s_________ 9_______145 m_____10 m/s_________ 10______150 m______5 m/s_________ 11______150 m______0 m/s_________ To work this problem, fill out the table given above with the accleration. Use the definition of average acceleration given earlier in this exercise and recognize that speed and velocity are the same for one dimensional motion. Average acceleration = ( velocity 2 - velocity 1 ) / total time = ( V2 - V1 ) / ( t2 - t1 ) Then answer the question below to check your answer. The next problem will give you the complete table so that you can check your answer. Question 4 options: 0 5 m/s^2 10 m/s^2 -5 m/s/s #5) Now let us look at the acceleration from the table. What is the average acceleration between 8 and 9 seconds? Time(s).|.Location(m).|.Velocity(m/s).|.Acceleration (m/s/s) 0_________0 m______0 m/s__________ 1_________5 m______5 m/s__________ 2________15 m_____10 m/s__________ 3________30 m_____15 m/s__________ 4________50 m_____20 m/s__________ 5________75 m_____25 m/s__________ 6_______100 m_____25 m/s__________ 7_______120 m_____20 m/s_________ 8_______135 m_____15 m/s_________ 9_______145 m_____10 m/s_________ 10______150 m______5 m/s_________ 11______150 m______0 m/s_________ To work this problem, fill out the table given above with the velocity and distance. Use the definition of average acceleration given earlier in this exercise and recognize that speed and velocity are the same for one dimensional motion. Average acceleration = ( velocity 2 - velocity 1 ) / total time = ( V2 - V1 ) / ( t2 - t1 ) Then answer the question below to check your answer. The next problem will give you the complete table so that you can check your answer. Question 5 options: 5 m/s^2 1 m/s^2 - 5 m/s^2 0 m/s^2 #6)Now let us look at the acceleration from the table. What is the average accleration between 8 and 9 seconds? Give your answer from the table you generated from the last question before you look at the table below. Time(s).|.Location(m).|.Velocity(m/s).|.Acceleration (m/s/s) 0_________0 m______0 m/s__________0 m/s/s 1_________5 m______5 m/s__________5 m/s/s 2________15 m_____10 m/s__________5 m/s/s 3________30 m_____15 m/s__________5 m/s/s 4________50 m_____20 m/s__________5 m/s/s 5________75 m_____25 m/s__________5 m/s/s 6_______100 m_____25 m/s__________0 m/s/s 7_______120 m_____20 m/s_________-5 m/s/s 8_______135 m_____15 m/s_________-5 m/s/s 9_______145 m_____10 m/s_________-5 m/s/s 10______150 m______5 m/s_________-5 m/s/s 11______150 m______0 m/s_________-5 m/s/s To work this problem, fill out the table given above with the and accleration. Use the definition of average acceleration given earlier in this exercise and recognize that speed and velocity are the same for one dimensional motion. Average acceleration = ( velocity 2 - velocity 1 ) / total time = ( V2 - V1 ) / ( t2 - t1 ) Question 6 options: 5 m/s/s 0 -5 m/s/s -10 m/s/s #7) We can say that the velocity of an object depends upon the acceleration the object has undergone but we cannot say that the acceleration is related to the velocity of the object. Let us apply this to a ball tossed upward. Velocity has to do with how fast a body travels. Acceleration has to do with how quickly a body changes the velocity. An upward thrown ball has an initial velocity that decreases with time. This rate of decrease is the acceleration, g, which is the same the entire time the ball is in the air. When the ball gets to the top of its trajectory, the instantaneous velocity is zero but the accleration is still g. As the ball returns, its velocity picks up at the same rate it decreased while rising (Ignoring air drag). Thus during the toss, the velocity changes but the acceleration does not. A ball is tossed vertically on the moon. It rises upward, reaches it highest point, and then falls back to its starting point. During this time, the acceleration of the ball is always Question 7 options: in the direction of the motion opposite the velocity of the ball directed upward directed downward #8) Drop a rock from a 5 meter height and it accelerates at 10m/s/s and strikes the ground 1 second later. Drop the same rock from a height of 2.5 m and its acceleration of fall is Question 8 options: about half the same more #9) If there is uniform acceleration, the final velocity is just the initial velocity plus the acceleration times the time of the acelleration. The final speed is the intital speed plus the gain of speed or the inital speed plus [(rate of gain of speed) times (the time under which the object gains speed)] Thus in algebraic form, if a is acceleration, t is time, and v0 is the initial velocity and v is the final velocity, then the speed at time t is just given by the relationship v = v0 + at An object fall freely from rest on a planet where the acceleration due to gravity is 20 m/s/s. After 5 seconds, the object will have a speed of Question 9 options: 5 m/s 10 m/s 20 m/s 50 m/s 100 m/s #10) From the change of velocity, we can determine the average acceleration. Since the final velocity is the initial velocity plus the acceleration times the time duration of the accleration, we can state the following. The average acceleration is the change of velocity over the the time it took to make the change. If a baseball being thrown goes from zero to 30 m/s in 0.1 seconds, what is its average acceleration? Question 10 options: 3 m/s/s 30 m/s/s 300 m/s/s 3600 m/s/s None of these

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plzzz split the question as its very cumbersome to read and confuses me as well as u thank u

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