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ACCELERATION

ACCELERATION. ACCELERATION. Vector quantity which measures how velocity changes over time Determined by the difference between the initial velocity (Vi) and final velocity (Vf) over change in time a ave = (v f – v i )/(t f –t i ) = D v/ D t

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ACCELERATION

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  1. ACCELERATION

  2. ACCELERATION Vector quantity which measures how velocity changes over time Determined by the difference between the initial velocity (Vi) and final velocity (Vf) over change in time a ave = (vf – vi)/(tf –ti) = Dv/Dt Represented by the unit distance over time2 (ex) 12 m/s2 45 km/h2

  3. Accelerating acceleration!?! For this class, we assume that all objects have constant acceleration. When an object accelerates, its displacement per unit time (velocity) changes.

  4. ACCELERATION EQUATIONS a = Vf – Vi Vf = Vi + at Δt Vf 2 = Vi2 + 2 a Δ d Δ d = Vi t + ½ a Δ t 2 Where a = acceleration Vf = final velocity Vi = initial velocity Δ d = change in displacement Δ t = change in time

  5. Vf = Vi + at Describes change in velocity under uniform acceleration. Tells how fast a particle will be going at time t if at time zero its velocity was vi. Converted to Vf = at if Vi = 0.

  6. Δ d = Vi t + ½ a Δ t 2 Describes change in position under uniform acceleration Also known as "equation of motion." Determines where a particle will be at time t, if at time zero velocity vi

  7. Vf2 = Vi2 + 2 a Δ d Tells what the final velocity (Vf) of the particle is given the initial velocity Vi , , acceleration rate and distance Δd Uses square root function to find Vf Simplifies to Vf2 = 2ad when vi = 0

  8. ACCELERATION PROBLEM Find average acceleration of a car that accelerates along straight road from rest to 80 km/h in 5 s . Vf = 80 km/h x 1h/3600s x 1000 m/km = 22 m/s Vi = 0 m/s a = (Vf – Vi)/ t = (22 m/s – 0 m/s)/5 s = 4.4 m/s 2

  9. OBJECTS THAT ARE SLOWING DOWN Has negative acceleration (deceleration) Find average acceleration of a car moving to the right(+x direction) 15.0 m/s when the driver brakes to 5.0 m/s in 5.0 s? a =Dv/Dt = (5.0 m/s – 15.0 m/s) ÷ 5.0 s = -2.0 m/s2 (acceleration negative) Would acceleration still be negative if car was moving to the left? NO! Its acceleration would be directed to the right and be positive.

  10. A shuttle bus slowed to a stop with an average acceleration of -1.8 m/s2. How long did it take the bus to slow down from 9.0m/s to 0.0 m/s? ` a = -1.8 m/s2 Vf = 0.0 m/s Vi = 9.0m/s Δ t = ? Δt = Vf – Vi = (0.0 m/s – 9.0 m/s) = -9.0 m/s a -1.8 m/s2 -1.8 m/s2 = 0.5 s

  11. A plane started from rest at one end of a runway underwent a uniform acceleration of 4.8 m/s2 for 15.0s before take off. What was its speed at takeoff? How long must the runway be for the plane to be able to take off? a = 4.8 m/s2Δ t = 15.0 s Vi = 0.0 m/s Vf = ? Vf = Vi + aΔt = 0.0 m/s + (4.8 m/s2)(15.0 s) = 72 m/s Δd = Vit + ½ a Δ t 2 = (0.0 m/s)(15.0 s) + ½ (4.8 m/s2) (15.0 s)2 = 540 m

  12. Final velocity after any displacement A person pushed a stroller from rest and uniformly accelerated at a rate of 0.500 m/s2. What was the velocity of the stroller after it has traveled 4.75 m? a = 0.500 m/s2 Δ d = 4.75 m Vi = 0.0 m/s Vf = ? Vf2 = Vi2 + 2 a Δ d = 0 m/s + 2 (0.500 m/s2 ) (4.75 m) Vf = 4.75 m/s

  13. Properties of acceleration In a velocity vs. time graph, the slope represents the object’s acceleration. In an acceleration vs. time graph, the area represents the object’s change in velocity.

  14. Is negative acceleration slowing down? Not always Remember the (+) and (-) just tell us the direction of the quantity

  15. Determining time with constant acceleration and distance given What is the time needed by a car to travel 30 m while accelerating at 2.00 m/s2 from rest. x = ½ a t2 t2 = 2x/a t = (2x/a)1/2 t = (2(30m)/2.00 m/s2) = 5.48s a = 2.00m/s2 xi = 0 vi = 0 xf = 30m

  16. Stopping distance A car traveling at 28 m/s decelerates at -6.0 m/s2. How many meters would it cover before coming to rest? Solution 1.Use vf2 = vi2+ 2a(xf-xi) Solve for xf = xi + (vf2 – vi2) ÷ 2a x = (0 – 28 m/s)2 ÷ 2(-6.0 m/s2) = 65m Solution 2.Use vav = (vf+ vi)/2 = 14 m/s vf = vi +at ; t = (vf – vi)/a = -28m/s ÷ -6.0 m/s2 = 4.67 s x = vav t = 14 m/s (4.67s) = 65 m

  17. Free Fall Which falls faster: a feather or a hammer?

  18. Galileo’s Experiment Which would reach the ground first, a marble or a cannonball?

  19. Galileo’s Discovery In the absence of air resistance, all objects undergo equal acceleration as it falls to the earth. The value of acceleration due to gravity (g) is equal to: g = 9.80 m/s2

  20. Maximum height of object thrown vertically upward A person throws ball upward with an initial velocity (viy ) of 15.0 m/s. (a) What is its maximum height (dymax)? (b) How long is it in the air? Since going up is positive, then vf2 = vi2 +2ay : at highest point, is vf = 0, then dymax = (vf2 – vi2)/2a = (0 – (15.0m/s)2)/2(-9.80m/s2) = 11.5 m

  21. Maximum Time Object is in the Air (1) Use dy = vit + ½ a t 0 = (15.0 m/s)t + ½(-9.80 m/s2)t2 Factor: (15.0 m/s – 4.90 m/s2t)t = 0 Since this is a quadratic equation, there are two solutions: t =0 corresponding to time ball was thrown upward t = 15.0 m/s ÷ 4.90 m/s2 = 3.06 s corresponds to instant ball returns to ground

  22. Maximum Time Object is in the Air (2) Using the time needed to reach highest point and equation: vf = vi +at t = (vf – vi)/a = (0 – 15 m/s)/-9.80 m/s2 = 1.53s Since this represents half the time object is up in the air, by symmetry: Total time = 2 x 1.53 s = 3.06 s

  23. Kinematics Graph Position vs time graph (x or y vs. t) Slope is velocity Velocity vs time graph Slope is acceleration Area under graph is displacement Acceleration vs time graph Constant and non zero for uniform acceleration

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