LT1: KInematics For 1D with solutions
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The document contains a physics test on kinematics in one dimension, including conceptual questions, problem solving questions, and an answer key. The conceptual questions cover topics like displacement, velocity, acceleration, and motion under gravity. The problem solving questions involve calculating values like speed, displacement, time and height using kinematic equations.
LT1: KInematics For 1D with solutions
- 1. 566547031750Questionnaire No.:_________00Questionnaire No.:_________8382000PHILIPPINE SCIENCE HIGH SCHOOL – Main Campus<br />Physics 2 – Advanced Topics in Physics I<br />1ST QUARTER - LONG TEST 1: KINEMATICS IN ONE DIMENSION<br />CONCEPTUAL QUESTIONSDirection:<br />Analyze carefully each question and determine the correct answer. Write your answer in your answer sheet.<br />(1 point) For which one of the following situations will the path length equal the magnitude of the displacement? Write only the letter of your choice.
- 2. A. Lydia de Vega is running around the PSHS oval.
- 3. B. A golf ball is rolling down an inclined plane.
- 4. C. A Honda CR-V travels 5 miles east; and then, it stops
- 6. D. An apple rises and falls after being thrown straight up
- 8. (4 points) A Toyota Fortuner, traveling at constant speed, makes two complete laps around a circular track of radius r in a time t. (a) When the Fortuner has traveled its one complete lap, what is the magnitude of its displacement? (b) What is the average speed of the car for two complete laps? (c) What is the average velocity’s magnitude for one complete lap? and (d) two complete laps?
- 9. (2 points) Joe is driving his car. He is initially driving to the right at constant speed. He sees the green traffic light changes to yellow, so he smoothly steps the brake. (a) Is the magnitude of its velocity greater than, less than, or equal to its initial velocity? What about the direction of its velocity? (b) Is his car accelerating? If yes, where is its acceleration?
- 11. (6 points) An object is initially moving to the left at constant speed from x=+2.00 m to x=0.00 m for 3.00 s. Then it stops for 1.00 s, reverses its direction and uniformly accelerates back to its initial position for 2.00 s. Sketch the graphs of (a) position-time, (b) velocity-time, and (c) acceleration-time.
- 12. (2 points) A ball is thrown vertically upward from the surface of the earth. (a) At what point in its path is its acceleration zero? (b) At what point in its path is its velocity zero?
- 13. PROBLEM SOLVINGDirection:<br />Solve all the problems completely and correctly in your answer sheet (Follow the format: Illustration: -> Given: -> Find: -> Derivation: -> Substitution: -> Final Answer: Box your final equation and answer. Follow significant digits rules.<br />(5 points) A car has an average speed of 11 m/s during the first 18.0 minutes (1080 s) of a 1.0-hour (3600 s) trip. What must the average speed of car during the last 42.0 minutes (2520 s) of the trip be if the car is to have an average speed of 21 m/s for the entire trip?
- 14. (5 points) Suppose an object’s velocity (moving in the same direction) was measured at five different moments as given by the following table:
- 15. Time, t (s)Velocity, v (m/s)0.00.001.01.002.02.003.03.004.03.00
- 16. Plot these points in a graph and compute the average velocity of its entire trip.
- 17. (10 points) A record of travel along a straight path is as follows:
- 18. Start from rest with a constant acceleration of 1.56 m/s2 for 14.0 s.
- 19. Maintain a constant velocity for the next 1.08 minutes.(a) What was the final velocity in leg 1 which is the velocity in leg 2? (b) What was the displacement for the trip?<br />(5 points) A ball is thrown vertically upward with a speed of 25.0 m/s. (a) How high does it rise? (b) How long does it take to reach its highest point?GOD Bless!!! John 3:16<br />Tract: sirdq_06/19/2011_2251h<br />PHILIPPINE SCIENCE HIGH SCHOOL – Main Campus<br />Physics 2 – Advanced Topics in Physics I<br />1ST QUARTER - LONG TEST 1: KINEMATICS IN ONE DIMENSION<br />ANSWER KEY<br />CONCEPTUAL QUESTIONS
- 20. (1 point) For which one of the following situations will the path length equal the magnitude of the displacement?
- 21. A. Lydia de Vega is running around the PSHS oval.
- 22. B. A golf ball is rolling down an inclined plane.
- 23. C. A Honda CR-V travels 5 miles east; and then, it stops and travel 2 miles west.
- 24. D. An apple rises and falls after being thrown straight up from the earth’s surface.
- 25. (4 points) A Toyota Fortuner, traveling at constant speed, makes two complete laps around a circular track of radius r in a time t. (a) When the Fortuner has traveled its one complete lap, what is the magnitude of its displacement? – 0
- 26. (b) What is the average speed of the car for two complete laps? – 4πrt
- 27. (c) What is the average velocity’s magnitude for one complete lap? and (d) two complete laps? – 0
- 28. (2 points) Joe is driving his car. He is initially driving to the right at constant speed. He sees the green traffic light changes to yellow, so he smoothly steps the brake.
- 29. (a) Is the magnitude of its velocity greater than, less than, or equal to its initial velocity? What about the direction of its velocity?
- 30. - less than, to the right
- 31. (b) Is his car accelerating? If yes, where is its acceleration? – to the left
- 33. (6 points) An object is initially moving to the left at constant speed from x=+2.00 m to x=0.00 m for 3.00 s. Then it stops for 1.00 s, reverses its direction and uniformly accelerates back to its initial position for 2.00 s. Sketch the graphs of (a) position-time, (b) velocity-time, and (c) acceleration-time.46672501530350024460201530350025654015303500<br />5733181144780<br />(2 points) A ball is thrown vertically upward from the surface of the earth.
- 34. (a) At what point in its path is its acceleration zero? – none (acceleration is due to gravitational force in free fall)
- 35. (b) At what point in its path is its velocity zero? – at its highest point/maximum height
- 36. PROBLEM SOLVING
- 37. (5 points) A car has an average speed of 11 m/s during the first 18.0 minutes (1080 s) of a 1.0-hour (3600 s) trip. What must the average speed of car during the last 42.0 minutes (2520 s) of the trip be if the car is to have an average speed of 21 m/s for the entire trip?
- 38. Illustration:
- 39. Given:v1=11 m/s- average speed for t1=1080 s
- 40. t2=2520 s- time to have the average speed v2
- 41. v=21 m/s - average speed for the entire trip at t=3600 sFind:v2=?- average speed for t2=2520 s<br />Derivation:v2=d2t2 where d2=d-d1for d1:d1=v1t1for d: d=vt<br />so that d2=vt-v1t1therefore: v2=vt-v1t1t2<br />Substitution:v2=21 m/s3600 s-11 m/s1080 s2520 sv2=25 m/s<br />(5 points) Suppose an object’s velocity (moving in the same direction) was measured at five different moments as given by the following table:
- 42. Time, t (s)Velocity, v (m/s)0.00.001.01.002.02.003.03.004.03.00
- 43. Plot these points in a graph and compute the average velocity of its entire trip. 1670053556000<br />Given: b=3.0 s - base of the right triangle in the graph (time interval in <br /> its uniform acceleration)<br /> h=3.00 m/s - height of the right triangle in the graph (change in <br /> velocity in its uniform acceleration)<br /> w=1.0 s - width of the rectangle in the graph (time interval in its <br /> uniform motion)<br /> <br /> L=3.00 m/s- length of the rectangle in the graph (velocity in its uniform motion)<br /> t=4.0 s- time for the entire trip<br />Derivation: Using the graph, the area under the line/curve is the displacement so<br /> d=12bh+Lwso, v=dt (valid to use in one-direction motion)v=12bh+Lwt<br />To simplify it further, the equation becomesv=bh+2Lw2t<br />Substitution:v=3.0 s3.00 m/s+23.00 m/s1.0 s24.0 sv=1.9 m/s,in the same direction<br />(10 points) A record of travel along a straight path is as follows:
- 44. Start from rest with a constant acceleration of 1.56 m/s2 for 14.0 s.
- 45. Maintain a constant velocity for the next 1.08 minutes.(a) What was the final velocity in leg 1 which is the velocity in leg 2? (b) What was the displacement for the trip?<br />Illustration:<br />Given:a=1.56 m/s2- acceleration in leg 1<br />vi=0- initial velocity in leg 1<br />t1=14.0 s- time interval in leg 1<br />t2=1.08 min=64.8 s- time interval in leg 2<br />a) Find:vf= ?- final velocity in leg 1<br />Derivation:In leg 1, it is in uniform acceleration so vf=vi+at1<br />since vi=0 the equation becomes vf=at1<br />Substitution:vf=1.56 m/s214.0 svf=21.8 m/s, in the same direction<br />b) Find:d= ?- displacement fo the entire trip<br />Derivation:In a straight path, the displacement’s magnitude is the same as the distance so d=d1+d2<br />Where d1=vit1+12at12(distance traveled in leg 1) which becomes d1=12at12<br />d2=vft2(distance traveled in leg /uniform velocity) <br />Since vf=at1 d2=at1t2 therefore d=12at12+at1t2 <br />Simplifying it furtherd=at112t1+t2<br />Substitution:d=1.56 m/s214.0 s1214.0 s+64.8 sd=1570 m, in the same direction<br />(5 points) A ball is thrown vertically upward with a speed of 25.0 m/s. (a) How high does it rise? (b) How long does it take to reach its highest point?
- 46. Illustration:
- 47. Given:vi=25.0 m/s- initial velocity of the ball thrown upward
- 48. a=g=-9.8 m/s2- acceleration due to gravity
- 49. vf=0- final velocity/velocity at its highest pointa) Find:d= ?- maximum height/highest point of the ball reached<br />Derivation:vf 2-vi 2=2adsince vf=0then -vi 2=2ad so d=-vi 22g<br />Substitution:d=-25.0 m/s22-9.8 m/s2d=31.9 m<br />b) Find:t= ?- time to reach its maximum height<br />Derivation:vf-vi=atsince vf=0then -vi=at so t=-vig<br />Substitution:d=-25.0 m/s-9.8 m/s2t=2.55 s<br />Sirdq_07/03/2011_0020h<br />