Lecture 5 : Introduction to Physics PHY101

1. Lecture 5: Introduction to PhysicsPHY101 Chapter 2: • Equations of Kinematics for Constant Acceleration in 1 Dim. (2.4, 2.5, 2.7) • Free Fall (2.6) Chapter 3: • Equations of Kinematics for Constant Acceleration in 2 Dim. (3.1, 3.2)

2. Summary of concepts from last lecture • position: your coordinates (just “x” in 1-D) • displacement: x = change of position • velocity: rate of change of position • average : x/t • instantaneous: slope of x vs. t : lim t->0x/t • acceleration: rate of change of velocity • average: v/t • instantaneous: slope of v vs. t : lim t->0v/t

3. v +x a  correct v +x  correct a Concept Question • A car is moving along the negative x direction. During part of the trip, the speed increases from 16 to 28 m/s in 1.5 s. Which of the following is correct? 1) v>0, a>0 2) v>0, a<0 3) v<0, a>0 4) v<0, a<0 • During another part of the trip, the speed decreases from 48 to 26 m/s in 11 s. Which of the following is correct? • 1) v>0, a>0 • 2) v>0, a<0 • 3) v<0, a>0 • 4) v<0, a<0 If speed is increasing, v and a are in same direction. If speed is decreasing, v and a are in opposite direction.

4. correct Concept Question Which of the following statements is most nearly correct? 1 - A car travels around a circular track with constant velocity. 2 - A car travels around a circular track with constant speed. 3- Both statements are equally correct. On a circular track, the direction that the car is traveling in is always changing, and since velocity takes into account the direction of travel, the velocity is always changing. Speed, however, is independent of direction and so the speed can stay constant.

5. Kinematics in One DimensionConstant Acceleration Consider an object which moves from the initial position x0, at time t0 with velocity v0, with constant acceleration along a straight line. How does displacement and velocity of this object change with time ? aav=a = (v-v0) / (t-t0) => v(t) = v0 + a (t-t0) (1) vav = (x-x0) / (t-t0) = (v+v0)/2 => x = x0 + (t-t0) (v+v0)/2 (2) Use Eq. (1) to replace v in Eq.(2): x(t) = x0 + (t-t0) v0 + a/2 (t-t0) 2 (3) Use Eq. (1) to replace (t-t0) in Eq.(2): v2 = v02 + 2 a (x-x0 ) (4)

6. Application of Eqs. of Kinematics • A runner accelerates to a velocity of 5.36 m/s due west in 3.00 s. His average velocity is 0.640 m/s2 due west. What was his velocity when he began accelerating ? [Chapter 2, problem #15] t0= 0 s, v= -5.36 m/s, t=3.00 s, aav=-0.640 m/s2 v0 = ? m/s aav = (v-v0)/(t-t0) => v0= v- aav (t-t0) = -3.44 m/s v0 = 3.44 m/s due west

7. Application of Eqs. of Kinematics • A drag racer starting from rest, speeds up for 402 m with a=+17 m/s2. A parachute then opens, slowing the car down with a=-6.10 m/s2. How fast is the racer after moving 3.50 x 102 m after the parachute opens ? [2-28] 1. Before the parachute opens (car moves +x direction): t0= 0 s, v01 = 0 m/s, x1=+402 m, a1=+17 m/s2 2. After the parachute opens: t0= 0 s, x2=+3.50 x 102 m, a2=-6.10 m/s2, v=? m/s v2=v022+2 a2 x2 Get v022 from 1.: v02=(2 a1 x1 )1/2=+117 m/s => v2=(v02+2 a 2 x2)1/2=+96.9 m/s

8. Free Fall • Free fall is the idealized description of the motion of a downward falling body due to gravity: • Air resistance is neglected • Acceleration due to gravity is considered to be constant The acceleration due to gravity is always pointing downward with magnitude g=9.80 m/s2.

9. Correct x = 1/2at2 Correct v=at Concept Question An object is dropped from rest. If it falls a distance D in time t then how far will it fall in a time 2t ? 1. D/42. D/23. D4. 2D5. 4D Followup question: If the object has speed v at time t then what is the speed at time 2t ? 1. v/42. v/23. v4. 2v5. 4v

10. correct Concept Question A ball is thrown vertically upward. At the very top of its trajectory, which of the following statements is true: 1. velocity is zero and acceleration is zero2. velocity is not zero and acceleration is zero3. velocity is zero and acceleration is not zero4. velocity is not zero and acceleration is not zero The velocity vector changes from moment to moment, buts its acceleration vector does not change. Though the velocity at the top is zero, the acceleration is still constant because the velocity is changing.

11. Free Fall • A wrecking ball is hanging from rest from a crane when suddenly the cable breaks. The time it takes the ball to fly half way to the ground is 1.2 s. Find the time for the ball to fall from rest all the way to the ground. [2-45] • Half way to the ground (-y direction) t0= 0 s, v0 = 0 m/s, t=1.2 s, a=-9.80 m/s2 Y1/2=v0 t + ½ a t2 = -7.1 m 2. From rest all the way to the ground, y=2 Y1/2 t0= 0 s, v0 = 0 m/s, a=-9.80 m/s2, t= ? s Y=v0 t + ½ a t2 = ½ a t2 => t= (2 y/a)1/2=1.7 s

12. Correct: v2 = v02 -2gy v0 Dennis Carmen v0 H vA vB Concept Question Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground? 1. Dennis' ball2. Carmen's ball3. Same

13. Kinematics in Two DimensionsConstant Acceleration Consider an object which moves in the (x,y) plane from the initial position r0, at time t0 with velocity v0, with constant acceleration. • position: your coordinates (just r=(x,y) in 2-D) • displacement: r = r-r0 change of position • velocity: rate of change of position • average : r/t • instantaneous: lim t->0r/t • acceleration: rate of change of velocity • average: v/t • instantaneous: lim t->0v/t Same concepts as in one dimension ! Equations of kinematics are derived for the x and y components separately. Same equations as in one dimension !