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- 2. UNIFORMLY ACCELERATED MOTION Uniformly Accelerated Motion is the motion of a point such that its tangential acceleration w is constant. In the case of uniformly accelerated rectilinear motion, the point's acceleration w is constant. The speed ? of the point t sec after uniform acceleration begins and the point's distance s from its initial position are determined for uniformly accelerated motion by the equations.
- 3. the motion of a point such that its tangential acceleration wτ is constant; in the case of uniformly accelerated rectilinear motion, the point’s acceleration w is constant. The speed ν of the point t sec after uniform acceleration begins and the point’s distance s from its initial position—s being measured along the point’s path— are determined for uniformly accelerated motion by the equations ν = ν0 + wτt s = v0t + wτt2/2 where ν0 is the initial speed of the point. When ν and wT are of the same sign, acceleration occurs; when they are of opposite sign, deceleration occurs.
- 4. When a rigid body undergoes uniformly accelerated translational motion, the above definitions apply to each point of the body. A body may also undergo uniformly accelerated rotation about a fixed axis; in this case, the body’s angular acceleration; ε is constant, and the angular speed ω and angular displacement ɸ of the body are ω = ω0 + ∊ t ɸ = ω0t t + ∊t2/2
- 6. PROBLEM NO.1 From rest, a car accelerated at 8 m/s2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds?
- 7. SOLUTION TO PROBLEM NO.1 a) The car starts from rest therefore the initial speed u = 0. Nothing is said about the initial position and we therefore assume it is equal to 0. Hence the position x is given by the equation x = (1/2) a t 2 where a is the acceleration (=8 m/s2) and t is the period of time between initial and final positions x = (1/2)8 (10)2 = 400 m b) The velocity v of the car at the end of the 10 seconds is given by v = a t = 8 * 10 = 80 m/s
- 8. PROBLEM NO. 2 With an initial velocity of 20 km/h, a car accelerated at 8 m/s2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds?
- 9. SOLUTION TO PROBLEM NO. 2
- 10. PROBLEM NO. 3 A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds. a) What is the acceleration of the car in m/s2? b) What is the position of the car by the time it reaches the velocity of 72 km/h?
- 11. SOLUTION TO PROBLEM NO. 3
- 12. PROBLEM NO. 4 Problem 4: An object is thrown straight down from the top of a building at a speed of 20 m/s. It hits the ground with a speed of 40 m/s. a) How high is the building? b) How long was the object in the air?
- 13. SOLUTION TO PROBLEM NO. 4
- 14. PROBLEM NO. 5 A train brakes from 40 m/s to a stop over a distance of 100 m. a) What is the acceleration of the train? b) How much time does it take the train to stop?
- 15. SOLUTION TO PROBLEM NO. 5a) We are given the initial velocity u = 40 m/s, the final velocity v = 0 (train stops) and the distance. Hence the formula that relates these 3 quantities and the acceleration is given by v2 = u2 + 2 a x 02 = 402 + 2 a (100) Solve for the acceleration a a = -1600 / 200 = - 8 m/s2 b) There two ways to find the time: 1) Use: x = (1/2)(v + u) t 100 = 0.5(0 + 40) t Solve for t: t = 5 seconds. 2) Use x = (1/2) a t2 + ut 100 = 0.5 ( - 8) t2 + 40t 4 t2 - 40 t + 100 = 0 4 (t2 - 10 t + 25) = 0 4(t - 5)2 = 0 t = 5 seconds.
- 16. PROBLEM NO. 6 A boy on a bicycle increases his velocity from 5 m/s to 20 m/s in 10 seconds. a) What is the acceleration of the bicycle? b) What distance was covered by the bicycle during the 10 seconds?
- 17. SOLUTION TO PROBLEM NO. 6
- 18. PROBLEM NO. 7 a) How long does it take an airplane to take off if it needs to reach a speed on the ground of 350 km/h over a distance of 600 meters (assume the plane starts from rest)? b) What is the acceleration of the airplane over the 600 meters?
- 19. SOLUTION TO PROBLEM NO. 7 a) In this problem the initial velocity u = 0 (assumed because it is not given) , the final velocity v = 350 km/h and the distance x = 600 meters = 0.6 km The relationship between the give quantities is: x = (1/2)(v + u) t 0.6 = 0.5 (350 + 0) t Solve for t t = (0.6 / 175) hours = 12.3 seconds b) The acceleration a of the airplane is given by a = (v - u) / t = 350 km/h / 12.3 s Convert 350 km/h into m/s 350 km/h = 350,000 m / 3,600 s = 97.2 m/s a = 97.2 m/s / 12.3 s = 8 m/s2 (to the nearest unit)
- 20. PROBLEM NO. 8 Starting from a distance of 20 meters to the left of the origin and at a velocity of 10 m/s, an object accelerates to the right of the origin for 5 seconds at 4 m/s2. What is the position of the object at the end of the 5 seconds of acceleration?
- 21. SOLUTION TO PROBLEM NO. 8 a) In this problem, we may consider that the direction of the object is the positive direction and the initial position x0 = -20 meters (to the left of the origin), the initial velocity u = 10 m/s, the acceleration a = 4 m/s2 and the time is t = 5 seconds. The position is given by x = (1/2) a t2 + u t + x0 = 0.5 * 4 * (5)2 + 10 * 5 - 20 = 80 meters to the right of the origin.
- 22. PROBLEM NO. 9 What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s2 to come to rest?
- 23. SOLUTION TO PROBLEM NO. 9 a) In this problem the initial velocity u = 360 km/h, the final velocity v = 0 (rest) and the acceleration a = -10 m/s2. The distance x can be calculated using the formula v2 = u2 + 2 a x Convert 360 km/h into m/s: 360 km/h = (360 000 m) /(3600 s) = 100 m/s x = ( v2 - u2 ) / (2 a) = (0 - 10,000) / (-20) = 500 meters
- 24. PROBLEM NO. 10 To approximate the height of a water well, Martha and John drop a heavy rock into the well. 8 seconds after the rock is dropped, they hear a splash caused by the impact of the rock on the water. What is the height of the well. (Speed of sound in air is 340 m/s).
- 25. SOLUTION TO PROBLEM NO. 10a) In this problem we have: 1) a rock was dropped down the well and is uniformly accelerated downward due to gravity. If h is the height of the well and t is the time taken by the rock to reach the bottom of the well, then we have h = (1/2)(9.8) t 2 2) After the splash, the sound travels up the well at a constant speed of 340 m/s. Again the same height h of the well is given by h = 340 *(8 - t) : 8 - t is the time taken for the sount to travel from bottom to top where the sound is heard. The above equations give: (1/2)(9.8) t2 = 340 *(8 - t) 4.9 t2 + 340 t - 2720 = 0 Solve for t, two solutions: t = 7.24 s and the second solution is negative and is not valid. The height h of the well is calculated using one of the above equations: h = 340 *(8 - t) = 340 *(8 - 7.24) = 257 meters (approximated to the the nearest meter)
- 26. PROBLEM NO. 11 A rock is thrown straight up and reaches a height of 10 m. a) How long was the rock in the air? b) What is the initial velocity of the rock?
- 27. SOLUTION TO PROBLEM NO. 11 a) In this problem the rock has an initial velocity u. When the rock reaches a height of 10 m, it returns down to earth and the the velocity v = 0 when x = 10 meters. Hence v = -9.8 t + u 0 = -9.8 t + u u = 9.8 t x = (1/2)(u + v) t 10 = 0.5 (9.8 t + 0) t = 4.9 t2 Solve for t: t = 1.42 seconds b) u = 9.8 t = 9.8 * 1.24 = 14 m/s
- 28. PROBLEM NO. 12 A car accelerates from rest at 1.0 m/s2 for 20.0 seconds along a straight road . It then moves at a constant speed for half an hour. It then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car.
- 29. SOLUTION TO PROBLEM NO. 12 a) The car goes through 3 stages: stage 1: acceleration a = 1, initial velocity = 0, t = 20 s. Hence the distance x is given by x = (1/2) a t2 = (1/2) (1) 202 = 200 meters stage 2: constant speed v is the speed at the end of stage 1. v = a t = 1 * 20 = 20 m/s x = v t = 20 m/s * (1/2 hour) = 20 m/s * 1800 s = 36,000 meters stage 3: deceleration to a stop, hence u = 20 m/s and v = 0 (stop) x = (1/2)(u + v) t = (1/2)(20 + 0) 30 = 300 meters
- 30. PROBLEM NO. 13 A car was travelling at a speed of 70km/h, the driver saw a rabbit on the road and slammed on the breaks. After 6.0 seconds the car came to a halt, how far did the car travel from the point where the brakes were first pressed to the point where the car stopped?
- 31. SOLUTION TO PROBLEM NO. 13 We are given: Vi = 70km/h = 19.4 m/s Vf = 0km/h t = 6s d = ? Our formula for distance is d = 1/2 ( Vf + Vi ) × t d = 1/2 ( 19.4 + 0 ) × 6 d = 58.332 The car stopped after 58 metres. This car has very bad brakes, as 58 metres is a very long distance to come to a halt.
- 32. PROBLEM NO. 14 Bill jogs at 6.0km/h, he then decides to accelerate into a light run. Bill accelerates at 0.030km/s² as he runs through a distance of 40m What is Bill's final speed?
- 33. SOLUTION TO PROBLEM NO. 14 First we convert our given information into a uniform set of magnitudes, metres and seconds. We are given: Vi = 6.0km/h = 1.6m/s Vf = ? d = 40m a = 0.003km/s² = 0.30m/s² Lets rearrange our uniform acceleration equations to eliminate the time, which we are not given. 2da = Vf² - Vi² Vf² = 2da + Vi² Vf² = 2(40)(0.3) + 1.6² Vf² = 26.6 Vf = 5.2m/s Vf = 18.5km/h After accelerating Bill ends up running at 18 km/h.
- 34. PROBLEM NO. 15 In performing a dismount off the high bar (3.048 meters above the ground), a gymnast's center of gravity reaches a height above the bar, of 2.5 m. Find: a. What was the gymnast's takeoff velocity? b. What was the velocity of the gymnast at landing?
- 35. SOLUTION TO PROBLEM NO. 15

- a) The car has an initial velocity of 20 km/h, therefore the initial speed u = 20 km/h. Nothing is said about the initial position and we therefore assume it is equal to 0. Hence the position x is given by the equation x = (1/2) a t 2 + u t where a is the acceleration (=8 m/s2) and t is period of time between initial and final positions and u is the initial velocity. Since the time is given in seconds, we need to convert 20 km/h into m/s as follows: u = 20 km/h = 20 * 1km1 hour 1000 m1 km 1 hour3600 seconds= 5.6 m/s We now have x = (1/2) (8) 102 + 5.6*10 = 456 m b) v = at + u = 8*10 + 5.6 = 85.6 m/s
- a) The acceleration a is a measure if the rate of change of the velocity within a period of time. Hence u = change in velocitychange in time= v - ut= 72 km/h - 011.5 secondsWe now convert 72 km/h into m/s u = 72 km/h = 72 * 1km1 hour 1000 m1 km 1 hour3600 seconds= 20 m/s We now calculate the acceleration a a = (20 m/s) / (11.5 s) = 1.74 m/s2 (approximetd) b) Two ways to find the position x: 1) x = (1/2)(v + u) t or 2) x = (1/2) a t 2 + u t 1) We first use: x = (1/2)(v + u) t = 0.5*(20 m/s + 0)*11.5 = 115 m 2) We now use: (1/2) a t2 + u t = 0.5*1.74*(11.5) 2 + 0*t = 115 m
- a) We consider that the direction from ground up is the positive direction of the falling object. We are given the initial (-20 m/s) and final velocities (-40 m/s); the minus sign was added to take into account the fact that the falling object is moving in the negative direction. We know the gravitational acceleration (g = - 9.8 m/s2) acting on the falling object and we are asked to find the height of the building. If we consider the position of the object as being x (wth x = 0 on the ground), then we may use the equation relating the initial and final velocities u and v, the acceleration a and the initial (x0 which the height of the building) and final (x, on the ground) positions as follows: v2 = u2 + 2 a (x - x0) (-40 m/s)2 = (-20 m/s)2 + 2 (-9.8 m/s0) (0 - x0) Solve the above for x0x0 = 1200 / 19.6 = 61.2 m b) x - x0 = (1/2)(u + v)t -61.2 = 0.5(-20 - 40)t t = 61.2 / 30 = 2.04 s
- a) In this problem the initial velocity u = 5 m/s and the final velocity v = 20 m/s. The acceleration a of the bicycle is the rate of change of the velocity and is given as follows a = v - ut= 20 m/s - 5 m/s10 seconds= 1.5 m/s2b) There are two ways to find the distance covered by the bicyle in t = 10 seconds. 1) x = (1/2)(v + u) t = 0.5 (20 + 5) 10 = 125 m 2) x = (1/2) a t2 + u t = 0.5 * 1.5 * 100 + 5 * 10 = 125 m