1. Chapter 2 & 3 Study Guide

2. 1. A racing car starting from a constant speed of 14.5 m/s accelerates at a rate of 6.25 m/sec2. What is the velocity of the car after it has traveled 755m? v o = 14.5 a = 6.25 d = 755 v = v 2 = v o 2 + 2ad v 2 = (14.5) 2 + 2(6.25)(755) v 2 = 9648 v = 98.2 m/s

3. 2. A stone is thrown from the top of a building with an initial velocity of 35.5 m/s straight upwards. The building is 62.0 m high, and the stone just misses the edge of the roof on its way down. Determine a) the time needed for the stone to reach its maximum height v o = 35.5 v = 0 a = -9.8 t = ? v – v o = at 0 – 35.5 = -9.8t T = 3.63 sec b) Determine it’s maximum height v o = 35.5 v = 0 a = -9.8 d = ? v = v o 2 + 2ad 0 2 = 35.5 2 + 2(-9.8)d d = 64.3 m

4. 2. A stone is thrown from the top of a building with an initial velocity of 35.5 m/s straight upwards. The building is 62.0 m high, and the stone just misses the edge of the roof on its way down. Determine c) The velocity and position of the stone at t = 6.5 sec. v o = 35.5 a = -9.8 t = 6.5 v – v o = at V = at + v o = (-9.8)(6.5) + 35.5 V = -28.2 m/sec d) The velocity of the stone just before it hits the ground. v 2 = v o 2 + 2ad V 2 = 35.5 2 + 2(-9.8)(-62) V = 49.8 m/sec

5. 3. One swimmer in a relay race has a 0.50s lead and is swimming at a constant speed of 4.00 m/s. He has 50.0m to swim before reaching the end of the pool. A second swimmer moves in the same direction as the leader. What constant speed must the second swimmer have in order to catch up to the leader at the end of the pool? Lead of 0.50 sec at 4 m/sec = 2m lead Time to finish… Second swimmer must swim 50 + 2 meters in 12.5 sec.

6. 4. A woman is reported to have fallen 44m from the 17 th floor of a building and to have landed on a metal ventilator box, which she crushed to a depth of 46 cm. She suffered only minor injuries. Neglecting air resistance, calculate a) The speed of the woman just before she collided with the box. V o = 0 a = 9.8 d = 44 v 2 = v o 2 + 2ad v 2 = 0 + 2(9.8)(44) V = 29.4 m/s b. Her acceleration while in contact with the box c. The time it took to crush the box V o = 29.4 V = 0 d = 0.46m v 2 = v o 2 + 2ad 0 = 29.4 2 + 2(a)(0.46) a = -940 m/s 2 d = ½ (v o + v)t 0.46 = ½ (29.4 + 0)t t = 0.0313 sec

7. 5. While exploring a cave, an explorer starts at the entrance and moves the following distances. She goes 75m North, 250m East, 125m at an angle 30 o North of East, and 150m South. Find her resultant displacement. N S WE 150 75 250 30 o 125 X

8. N S WE 150 75 250 30 o 125 X 125 cos 30 125 sin 30 Σ Fx = 250 + 125 Cos 30 = 358 Σ Fy = 75 + 125 sin 30 – 150 = -12.5 5. While exploring a cave, an explorer starts at the entrance and moves the following distances. She goes 75m North, 250m East, 125m at an angle 30 o North of East, and 150m South. Find her resultant displacement.

9. N S WE 150 75 250 30 o 125 X 125 cos 30 75 sin 30 Σ Fx = 250 + 125 Cos 30 = 358 Σ Fy = 75 + 125 sin 30 – 150 = -12.5 358 12.5R R = 358 5. While exploring a cave, an explorer starts at the entrance and moves the following distances. She goes 75m North, 250m East, 125m at an angle 30 o North of East, and 150m South. Find her resultant displacement.

10. N S WE 150 75 250 30 o 125 X 125 cos 30 75 sin 30 Σ Fx = 250 + 125 Cos 30 = 358 Σ Fy = 75 + 125 sin 30 – 150 = -12.5 358 12.5R θ 5. While exploring a cave, an explorer starts at the entrance and moves the following distances. She goes 75m North, 250m East, 125m at an angle 30 o North of East, and 150m South. Find her resultant displacement.

11. 135m R Find time of flight… v o = 0 d = 135 a = 9.8 d = v o t + ½ at 2 135 = ½ (9.8)t 2 t = 5.25 sec R = v x ● t = (55)(5.25)= 289 m V 2 = v o 2 + 2ad V 2 = 0 + 2(9.8)(135) V y = 51.4 m/s V x = 55.0 m/s 6. A rescue plane drops a package of emergency rations to a stranded party of explorers. If the plane is traveling at 55 m/s at a height of 135 m above the ground… a)Where does the package strike the ground relative to the point at which it was released? b)What are the horizontal and vertical components of the impact?

12. 7. A long jumper leaves the ground at an angle of 22 o to the horizontal and at a speed of 12.5 m/s. A) How far does he jump? B) What is the maximum height reached C) how long is he in the air? Consider motion on the y axis… V o = 12.5 sin 22 o = 4.68 a = -9.8 d = 0 d = v o t + ½ at 2 0 = 4.68t – ½ (9.8)t 2 t = 0.955 sec

13. 8. A hockey player is standing on his skates on a frozen pond when an opposing player moving with a uniform speed of 12 m/s skates by with the puck. After 3.0 s the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.0 m/s 2 how long does it take him to catch his opponent? 36m12 m/s 4 m/s 2 d 1 = 36 + vt d 2 = ½ at 2 d 1 = d 2 when they meet at a time t 36 + 12t = ½ (4)t 2 2t 2 -12t – 36 = 0 T = 8.2 sec