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Solving Accelerated Motion Problems
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Solving Accelerated Motion Problems

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Shows step by step how to solve typical accelerated motion problems in physics. …

Shows step by step how to solve typical accelerated motion problems in physics.
**More good stuff available at:
www.wsautter.com
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  • 1. Solving Accelerated Motion Problems & (General Problem Solving Techniques) Copyright Sautter 2003
  • 2. The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. [email_address] More stuff at: www. wsautter .com
  • 3. Books available at: www. wsautter .com www.smashwords.com www.amazon.com www.bibliotastic.com www.goodreads.com Walt’s Books for Free!
  • 4. General Problem Solving Steps
    • ( 1) Read the problem more than once (three of four times is preferable)
    • (2) Decide what is to be found. Identify the algebraic symbol for the unknown (v i for instantaneous velocity, t for time, s i for instantaneous displacement, etc.)
    • (3) Identify the given information and the algebraic symbols which represent those values.
    • (4) Make a sketch of the situation which describes the problem (draw a picture when possible).
    • (5) Select an equation which contains the unknown value and the given values.
  • 5. General Problem Solving Steps
    • ( 6) If all the required values for an equation are not given find an alternative equation which allows for the calculation to values required for the unknown containing equation.
    • (7) Convert all given values to an appropriate measurement system (all MKS, CGS or English units)
    • (8) Solve the equation algebraically for the unknown value.
    • (9) Insert the numerical values in the equation and do the math.
    • (10) Does the answer make sense? For example if the problem requires us to find the time needed for an object to fall 100 feet is calculated as 20 minutes, this is obviously incorrect and we must try the problem again!
  • 6. Displacement vs time Velocity vs time slope Acceleration vs. time slope Area under curve Area under curve Displacement vs time derivative Velocity vs time Acceleration vs. time derivative integral integral Displacement, Velocity & Acceleration
  • 7. Accelerated Motion Equations
    • V AVERAGE =  s/  t = (V 2 + V 1 ) / 2
    • V INST. = V ORIGINAL + at
    • S INST = V 0 t + ½ at 2
    • S i = ½ ( Vi 2 – V o 2 ) / a
    • dy/dx = nx n-1
    •  dy =  x n dx = ( x n+1 / n+1) + C
    • Instantaneous velocity can be found by taking the slope of a tangent line at a point on a displacement vs. time graph.
    • Instantaneous velocity can also be determined from an acceleration vs. time graph by determining the area under the curve.
    • Displacement is given by the area under a velocity vs. time graph.
    • The slope of an velocity versus time curve is the acceleration.
  • 8. A car has an acceleration of 8.0 m/s 2 . (a) How much time is needed to reach a velocity of 24 m/s form rest? (b) How far does the car travel ?
    • Solution: part (a)
    • We are asked to find time (t). The original velocity is 0 (at rest is physics for zero velocity) (V o = 0). The velocity at the time to be found is 24 m/s (V i = 24 m/s). The acceleration is 8.0 m/s 2 (a = 8 m/s 2 )
    • The equation Vi = Vo + at contain all given terms and the term we want to find.
    • Solving for t gives t = (V i – V o )/a
    • t = (24 – 0) / 8 = 3.0 sec (MKS units are used)
    • Part (b) Find displacement (Si) at 3.0 seconds
    • Now that time is known we can use Si = V 0 t + ½ at 2
    • Si = (0 x 3.0) + ½ (8.0)(3.0)2 = 36 meters (MKS units)
    V o = 0 V i = 24 m/s a = 8.0 m/s 2 S i = ?
  • 9. RELATING SIGNS OF VELOCITY & ACCELERATION TO MOTION
    • VELOCITY ACCELERATION MOTION
      • + + MOVING FORWARD (UP) & SPEEDING UP
      • + - MOVING FORWARD (UP) & SLOWING
      • - - MOVING BACKWARD (DOWN) & SPEEDING UP
      • - + MOVING BACKWARD (DOWN) & SLOWING
      • + OR - 0 CONSTANT SPEED
      • 0 + OR - MOVING FROM REST
  • 10. The brakes of a car can give an acceleration of 6.0 ft/sec 2. (a) How long will it take to stop from a speed of 30 mph? (b) How far will it travel during the stopping time?
    • Solution: part (a) We are to find time (t). We are given the original velocity (V o ) but it is in miles per hour. We need feet per second to be consistent with the given acceleration units (ft/sec 2 ).
    • 1 mile = 5280 feet and 1 hour is 60 minutes x 60 seconds = 3600 seconds, therefore 30 mph x 5280 = 158,400 ft / 3600 sec = 44 ft/s. V 0 = 44 ft/s. When stopped, V i = 0 ft/s.
    • The car is slowing will moving forward therefore acceleration is negative and a = - 6.0 ft/s 2
    • An equation containing V o , V i , a and t is Vi = Vo + at . Solving for t we get t = (V i – V o )/a
    • t = (0 – 44)/- 6.0 = 5.0 sec
    • Now that time is known we can use Si = V 0 t + ½ at 2
    • t = (44 x 5.0) + ½ (- 6.0)(5.0) 2 = 145 feet
    V i = 0 V o = 44 ft/s a = - 6.0 ft/s 2 S = ?
  • 11. A plane needs a speed of 50 m/s in order to take off. The runway length is 500 meters. What must be the plane’s acceleration?
    • Solution: We will find acceleration (a). The plane is originally at rest (V 0 = 0). The velocity when the plane takes off is 50 m/s (V i = 50 m/s). The displacement of the plane is 500 m (S i = 500).
    • Since no time is given we will use S i = ½ (Vi 2 – V o 2 ) /a. It contains V i , V o , S i and a for which we are looking! All units are MKS so no conversions are needed.
    • Solving for a we get a = ½ (Vi 2 – V o 2 ) /S i
    • a = ½ (50 2 – 0 2 ) / 500 = 2.5 m /s 2
    V o = 0 V i = 50 m/s S i = 500 m a = ?
  • 12. Acceleration Due to Gravity & Free Fall CLICK HERE g = 9.8 meters / second 2 g = 980 centimeters / second 2 g = 32 feet / second 2
  • 13. A stone is dropped from a cliff. (a) What is its acceleration? (b) What is its speed after 3.0 seconds? ( c) How far has it fallen?
    • Solution (a) All objects under the sole influence of gravity are said to be in “free fall” and their acceleration is gravity (- 9.8 m/s 2 ) The negative sign means downward.
    • (b) The original velocity is 0. It is released from rest. The equation Vi = Vo + at contains all the given variables. (a = gravity). Vi = 0 + (- 9.8)(3.0) = - 29.4 m/s (negative means the object is moving downward) CGS or English unit can also be used depending on what gravity value is selected.
    • ( c) We are finding S i . The equation S i = V 0 t + ½ at 2 contains all our given variables and again acceleration = gravity.
    • Si = (0 x 3.0) + ½ (- 9.8)(3.0) 2 = - 44.1 meters (negative means below the point of release)
    V o = 0 V i = ? t = 3.0 s S i = ? a = g = - 9.8 m/s 2
  • 14. A stone is thrown downward from a cliff at 10 ft/s. (a) What is its acceleration? (b) What is its speed after 3.0 seconds and ( c) how far has it fallen?
    • Solution (a) Objects which are thrown, once they are released are under the sole influence of gravity and their acceleration is gravity (- 9.8 m/s 2 )
    • (b) The object is not released from rest. Vo = - 10 ft/s (thrown downward and thus negative). The equation Vi = Vo + at contains all the given variables. (a = gravity).
    • Vi = -10 + (- 32)(3.0) = - 106 ft/s (negative means the object is moving downward) English units are used
    • ( c) We are finding S i . The equation Si = V 0 t + ½ at 2 contains all our given variables and again acceleration = gravity.
    • Si = (-10 x 3.0) + ½ (- 32)(3.0) 2 = -174 feet (negative means below the point of release)
    V o = -10 ft/s V i = ? t = 3.0 s S i = ? a = g = - 32 ft/s 2
  • 15. WHEN AN OBJECT IS THROWN UPWARDS, ITS VELOCITY CHANGES DURING FLIGHT AS SHOWN BY THE VECTOR ARROW VERTICAL MOTION & GRAVITY GRAVITY = - 9.8 M / S 2 - 980 cm / S 2 - 32 ft / S 2 V i = 0 at the highest point of the flight
  • 16. A stone is thrown upward form the edge of a cliff at 10 ft/s. (a) What is its acceleration? (b) What is its speed after 3.0 seconds and ( c) how high is the object? V o = +10 ft/s V i = ? t = 3.0 s S i = ? a = g = - 32 ft/s 2
    • Solution (a) Objects which are thrown, once they are released are under the sole influence of gravity and their acceleration is gravity (- 9.8 m/s 2 )
    • (b) The object is not released from rest. Vo = + 10 ft/s (thrown upward and thus positive). The equation Vi = Vo + at contains all the given variables. (a = gravity).
    • Vi = +10 + (- 32)(3.0) = - 86 ft/s (negative means the object is moving downward) English units are used
    • ( c) We are finding S i . The equation Si = V 0 t + ½ at 2 contains all our given variables and again acceleration = gravity.
    • Si = (+10 x 3.0) + ½ (- 32)(3.0) 2 = - 114 feet (negative means below the point of release – the of the cliff ! )
  • 17. A stone is thrown upward form the edge of a cliff at 30 ft/s. (a) What is its maximum height? (b) What is its speed when it is 30 ft below the edge of the cliff? s V o = +30 ft/s V i = 0 (a) S i = -50 ft (b) a = g = - 32 ft/s 2
    • Solution (a) At the highest point V i = 0. V o = + 30 ft/s and a = g = - 32 ft/s 2 . The equation we will use is S i = ½ (Vi 2 – V o 2 ) /a
    • S i = ½ (0 2 – (+30) 2 ) / - 32 = + 14.1 feet above the point of release (the edge of the cliff)
    • (b) The object is 50 feet below the point of release and therefore S i = - 50 ft. Again, we will use S i = ½ (Vi 2 – V o 2 ) /a
    • Solving for V i we get V i = ((2 x S i x a ) + V o 2 ) ½
    • V i = ((2 x (-50) x (-32)) + (+30) 2 ) ½ = + or – 64 ft/s (all square root values for positive numbers are + or -). Since the object is below the point of release it must be falling and – 64 ft/s is the correct answer !
    S i = ? (a) V i = ? (b)
  • 18. A ball is thrown vertically into the air at 10 m/s. How long will it take to land?
    • There are two ways to solve this problem
    • Method 1 : At the highest point the velocity V i =0. We will use Vi = Vo + at. Solving for t we get t = (V i – V o )/a and t = (0 – 10) / -9.8 = 1.02 seconds to the high point and of course an equal time (1.02 sec) to fall back to the ground. Total time then is 2 x 1.02 = 2.04 seconds.
    • Method 2 : When the ball is back on the ground its vertical height is 0. We will use S i = V 0 t + ½ at 2 . Then 0 = 10 t + ½ (-9.8) t 2 or 4.9 t 2 –10t = 0. Dividing both sides of the equation by t we get 4.9 t – 10 = 0 and solving for t gives 2.04 seconds (the same answer as method 1)
    V i = 0 V o = +10m/s t = ? a = g = - 32 ft/s 2
  • 19. Now it's time for you to try some problems on your own ! Each problem is similar to one of the solved problems so look back and review the appropriate problem if you get stuck ! Don't forget to use the steps outlined at the beginning of this program.
  • 20.
    • A ball is thrown upwards at 12 m/s.
    • How high will it be in 1.0 seconds?
    • (A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m
    Click Here For answers What is its maximum height ? (A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m What is its height in 2.0 seconds ? (A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m What is its acceleration ? (A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m NONE
    • Gravity is the
    • acceleration
    • It is
    • 9.8 m / s 2
    • (MKS units)
    What is its acceleration at the highest point when the velocity is zero ? (A) –9.8 m/s 2 (B) –980 cm/s 2 (C) –32 ft/s 2 They are All right ! Gravity never Turns off !
  • 21. A pitch is thrown at 80 mph. It takes 0.1 seconds to throw the ball. What is the acceleration? (A) 1176 ft/s 2 (B) 32 ft/s 2 (C) 800 ft/s 2 (D) 2560 ft/s 2 A car comes to stop from 60 m/s over a distance of 360 meters. What is the car’s acceleration ? (A) -10 ft/s 2 (B) – 5.0 ft/s 2 (C) –2.2 ft/s 2 (D) – 0.50 ft/s 2 A ship has an average velocity of 30 mph. How Far does it travel in day? (A) 1058 miles (B) 1800 miles (C) 720 miles (D) 1.25 miles
    • A boy throw a ball upward to a height of 20 ft. How
    • Long must he wait to catch it?
    • 0.625 sec (B) 1.00 sec (C) 2.24 sec (D) 1.12 sec
    A penny is dropped from the Sears Tower (1450 ft). What is its velocity when it hits the ground ? (A) –32 ft/s (B) –169 ft/s ( C) –305 ft/s (D) –201 ft/s Click Here For answers
  • 22. t = 0.1 sec 3 feet
    • From what height above
    • The top of the window
    • was the ball dropped ?
    • 12.6 ft (B) 15.0 ft
    • (C) 30 ft (D) 18.8 ft
    SOLUTION: The average velocity from top to bottom of the window  s/  t = (V 2 + V 1 ) / 2, ( V 2 + V 1 ) /2 = 3.0 / .1 = 30, V 2 + V 1 = 60, Vi = Vo + at , V 1 = 0 + 32 t 1 , V 2 = 0 + 32 t2, 32 t 2 + 32 t 1 = 60, t 2 + t 1 = 60/32 = 1.875 and t 2 - t 1 = 0.1, t 2 = 0.1 + t 1 , 0.1 + t 1 + t 1 = 1.875 t 1 = .8875 and using S i = V 0 t + ½ at 2 S i = (0 x 0.8875) + ½ (32)(0.8875) 2 = 12.6 feet Time to fall Passed window opening Now for a final Hard Problem! WATCH THE WINDOW Click Here For answers
  • 23. The End