Solving Accelerated Motion Problems
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The document outlines a systematic approach to problem-solving in physics, detailing steps from understanding the problem to interpreting the solution. It includes various equations related to motion and acceleration, and provides examples illustrating their application. Additionally, it discusses the relationships between velocity, acceleration, and motion, and presents practical problems involving calculations of time, displacement, and acceleration.
Solving Accelerated Motion Problems
- 2. 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 (vi for instantaneous velocity, t for time, si 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. 2
- 3. 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! 3
- 5. Accelerated Motion Equations • VAVERAGE = s/ t = (V2 + V1) / 2 • VINST. = VORIGINAL + at • SINST = V0 t + ½ at2 • Si = ½ (Vi 2 – Vo 2) /a • dy/dx = nx n-1 • dy = xn 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. 5
- 6. A car has an acceleration of 8.0 m/s2. (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) (Vo = 0). The velocity at the time to be found is 24 m/s (Vi = 24 m/s). The acceleration is 8.0 m/s2 (a = 8 m/s2) • The equation Vi = Vo + at contain all given terms and the term we want to find. • Solving for t gives t = (Vi – Vo)/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 = V0 t + ½ at2 • Si = (0 x 3.0) + ½ (8.0)(3.0)2 = 36 meters (MKS units) Vo = 0 Vi = 24 m/sa = 8.0 m/s2 Si = ? 6
- 7. 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 7
- 8. The brakes of a car can give an acceleration of 6.0 ft/sec2. (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 (Vo) but it is in miles per hour. We need feet per second to be consistent with the given acceleration units (ft/sec2). • 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. V0 = 44 ft/s. When stopped, Vi = 0 ft/s. • The car is slowing will moving forward therefore acceleration is negative and a = - 6.0 ft/s2 • An equation containing Vo , Vi , a and t is Vi = Vo + at . Solving for t we get t = (Vi – Vo)/a • t = (0 – 44)/- 6.0 = 5.0 sec • Now that time is known we can use Si = V0 t + ½ at2 • t = (44 x 5.0) + ½ (- 6.0)(5.0)2 = 145 feet Vi = 0Vo = 44 ft/s a = - 6.0 ft/s2 S = ? 8
- 9. 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 (V0 = 0). The velocity when the plane takes off is 50 m/s (Vi = 50 m/s). The displacement of the plane is 500 m (Si = 500). • Since no time is given we will use Si = ½ (Vi 2 – Vo 2) /a. It contains Vi , Vo , Si 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 – Vo 2) /Si • a = ½ (50 2 – 02) / 500 = 2.5 m /s2 Vo = 0 Vi = 50 m/s Si = 500 m a = ? 9
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