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Info. First Exam Thursday Oct. 2 Chapters 1, 2, and 3 homework due at that time. 3inch x 5 inch card is okay to use 1-side of equations, no sentences 1st lab due Monday Oct. 6 2nd lab Due Monday Oct. 13. Kinematic Description of Motion. Speed Distance Position Displacement Velocity

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**Info**First Exam Thursday Oct. 2 Chapters 1, 2, and 3 homework due at that time. 3inch x 5 inch card is okay to use 1-side of equations, no sentences 1st lab due Monday Oct. 6 2nd lab Due Monday Oct. 13 .**Kinematic Description of Motion**Speed Distance Position Displacement Velocity Acceleration Time**Kinematic Description of Motion**Speed Distance Time all scalar quantities Distance traveled Average speed = Time taken d Avg. sp = (m/s) t**Average Speed**Speed Distance Time all scalar quantities Distance traveled Average speed = Time taken d Avg. sp = (m/s) t**Instantaneous Speed**Speed at an instant in time How long does and instant last? Distance traveled Instantaneous speed = tiny time interval d (m/s) sp = ∆t**What is the average speed of a runner who runs 100.0 m in**11.0 sec? d Avg. sp = t Given: d=100.0 m t=11.0 s Want: avg. sp Use: 100.0 m = 9.09 m/s 11.0 s**A race car driver must average 100 mi/hr for a 200 mi trial?**In the first 100 miles the driver only averages 50 mi/hr. With what average speed must she travel the last 100 miles in order to qualify for the big race?**Displacement & Velocity**Displacement velocity Both vectors (Magnitude and direction) Displacement: The straight line distance between two points, along with the direction from starting point to final point.**Displacement & Velocity**Displacement (∆x) : The straight line distance between two points, along with the direction from starting point to final point. X1 position of point 1 , X2 position of point 2 x1 ∆x x2 ∆x=x2-x1 Displacement: Change in position**Displacement & Velocity**Displacement velocity Both vectors displacement Average velocity = Time taken ∆x V = (m/s), and direction t**Displacement & Velocity**What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs? Given ∆x=160 miles North, t=4.0 hr Want: V Use:**Displacement & Velocity**What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs? Given ∆x=160 miles North, t=4.0 hr Want: V Use: displacement Average velocity = Time taken ∆x 160 miles, N V = = = 40 mi/hr, N t 4.00 hr**Displacement & Velocity**What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs? Given ∆x=0 miles North, t=7.0 hr Want: V**Displacement & Velocity**What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs? Given ∆x=0 miles North, t=7.0 hr Want: V Use: displacement Average velocity = Time taken**Displacement & Velocity**What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs? Given ∆x=0 miles North, t=7.0 hr Want: V Use: displacement Average velocity = Time taken ∆x 0 miles, N V = = = 0 mi/hr, N t 7.00 hr**Displacement & Velocity**What is the average speed of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs? Given d=320 miles, t=7.0 hr Want: V**Displacement & Velocity**What is the average speed of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs? Given d=320 miles, t=7.0 hr Want: V Use: distance Avg. sp = Time taken**Displacement & Velocity**What is the average speed of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs? Given d=320 miles, t=7.0 hr Want: V Use: distance Avg. sp = Time taken d 320 miles Avg. sp = = = 45.7 mi/hr 7.00 hr t**Displacement & Velocity**X position at time t Xo initial position at t=0 (usually set to zero) average velocity**QUICK QUIZ 2.1**A car with a dripping gas tank travels along a road in a straight line. A drop falls from the gas tank every five seconds and over a section of road the spots from the gasoline can be used to determine the average speed of the car. From the spots in the figure below, the average speed of the car is determined to be a) 20 m/s, b) 24 m/s, c) 30 m/s, or d) 120 m/s?**QUICK QUIZ 2.1 ANSWER**(b). The average speed is the total distance divided by the total time elapsed. The distance between two successive spots represents a time interval of 5 seconds. There are therefore 5 time segments of 5 seconds so that the total time elapsed is 25 seconds. So vavg = Dx/Dt = 600 m/25 s = 24 m/s. It would be incorrect to simply count the number of spots and obtain a time equal to 6 X 5 = 30 seconds. Recall that it is the time interval, Dt, rather than the time, t, that belongs in the average velocity equation.**A car travels from Vancouverwith an average speed of 50**mi/hr. What can be said about where this car is after 1 hr? A car travels from Vancouverwith an average velocity of 50 mi/hr North for one hour . What can be said about where this car is after 1 hr?**A car travels with an average speed of 50 mi/hr from**Vancouver. What can be said about where this car is after 1 hr? It is somewhere wiithin or on the boundaries of a circle of radius 50 miles centered about vancouver. A car travels from Vancouver with an average velocity of 50 mi/hr North for one hour . What can be said about where this car is after 1 hr? It is 50 miles north of vancouver.**Average acceleration**Vxf velocity at time t Vxi intial velocity at t=0.**Average acceleration**What is the average acceleration of a jet that starts from rest and speeds up to 200 mi/hr in 8.0 sec?**Average acceleration**What is the average acceleration of a jet that starts from rest and speeds up to 200 mi/hr in 8.0 sec? Given: Vo=0.0 (rest) V=200 mi/hr t=8.0 s Want: a Use:**Average acceleration**Given: Vo=0.0 (rest) V=200 mi/hr t=8.0 s Want: a Use:**Average acceleration**What is the average acceleration of a bird that starts from rest and speeds up to 20.0 m/s in 8.00 sec? Given: Vxi=0.0 (rest), Vxf=20.0 m/s, t=8.0 s Want: ax Use:**Average acceleration**Given: Vo=0.0 (rest) V=20.0 m/s t=8.0 s Want: a Use:**Is it possible to accelerate while traveling at a constant**speed?**QUICK QUIZ 2.2**A dragster starts from rest and takes off down a race track. At the precise moment that the dragster starts to move, its initial velocity is zero. At this precise moment, the instantaneous acceleration of the dragster is a) in the same direction as the subsequent motion of the dragster, b) in the opposite direction to the subsequent motion of the dragster, or c) zero?**QUICK QUIZ 2.2 ANSWER**(a). The instantaneous acceleration is the derivative of the velocity with respect to time. Examining a possible graph of velocity versus time for the dragster reveals that, although the velocity is zero at t = 0, the derivative of the velocity (or slope of the graph) is not zero and is positive, indicating that the acceleration is non-zero and in the direction of the motion.**QUICK QUIZ 2.3**When a car skids after applying its brakes, the acceleration of the car is in the opposite direction to the velocity of the car. If the acceleration of the car would remain constant in this direction, the car would a) eventually stop and remain stopped, (b) eventually stop and then start to speed up in the forward direction, (c) eventually stop and then start to speed up in the reverse direction, or (d) never stop but continue to speed up in the forward direction? (end of section 2.4)**QUICK QUIZ 2.3 ANSWER**(c). If the acceleration would remain constant in a direction opposite the initial velocity direction, eventually the car would stop and then speed up in the reverse direction. A way to achieve such a feat would be to slam the car into reverse while traveling forward. The car would at first skid, then stop, and then speed up in reverse. For a constant reverse acceleration, the velocity curve is graphed below where it can be seen that the car instantaneously stops when the velocity is zero and then continues with a negative velocity.**A car traveling with a velocity of +30.0 m/s accelerates at**a rate of -4.0 m/s2. How long will it take this car to stop? Given vo = +30.0 m/s, a=-4.0 m/s2 and v = 0.0 Want: t Use:**Kinematic equations**Distance=Avg Sp (time) 1 2 3 Constant a 4 5**V vs t Graph and average Velocity**=Vxi Constant Velocity ax=__?___ V DX=Vxit = “AREA” Vo t**V vs t Graph constant acceleration**A straight line V Vo t**1**2 Constant a 3**Eliminate v from 1,2, &3**(use 3) (use 2) algebra 4

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