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Lecture 9

Lecture 9. “Instantaneous” velocity. If the velocity can change as time goes on, then we should be able to speak of the velocity at a particular instant of time. The average velocity is measured over a finite time interval ∆t.

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Lecture 9

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  1. Lecture 9

  2. “Instantaneous” velocity • If the velocity can change as time goes on, then we should be able to speak of the velocity at a particular instant of time. • The average velocity is measured over a finite time interval ∆t. • The instantaneous velocity is in principle the average velocity measured over an infinitesimal time interval dt.

  3. t 0 1 2 3 4 5 6 7 8 9 10 y 0 4.9 20 44 78 123 176 240 314 397 490 dy/dt 0 9.8 20 29 39 49 59 69 78 88 98 600 500 400 300 200 100 0 1 2 3 4 5 6 7 8 9 10 11 Example • An object in free frictionless fall can be modelled by an equation y = 4.9t2, where y=0 is the point where the object was released & down is the +ve y direction. What is the velocity • at time t=4 s? • at time t=8 s? • at time t=12 s? distance velocity v=dy/dt = 2x 4.9 t

  4. Acceleration • Rate at which velocity changes with time. • Average acceleration: • Instantaneous acceleration:

  5. Acceleration e.g. A car can go from rest to 100 km/h in 10 seconds. What is its average acceleration? • Use, average acceleration: 100000/60x60 10 =2.8m/s e.g. An object has position given by y = 4.9t2. What is its acceleration at t =1? d2 y/dt2 = 2x4.9 = 9.8m/s2

  6. 1-D kinematic models • Constant velocity is easy to deal with: • x = xi + vt (equivalently ∆x = v∆t) • Next step up is dealing with velocity which is changing uniformly, i.e. constant acceleration v=vi + at (or ∆v = a∆t) a= ∆v/ ∆t

  7. Kinematic Model (1-D, const a) • Constant velocity: ∆x = v∆t • Constant acceleration: ∆x = v0t + 1/2at 2 v ∆t Position, ∆x = Area under v vs. t graph t0 t ∆x = Area 1 + Area 2 = v0 ∆ t + ½ ∆ v ∆t We know a= ∆ v/ ∆t; so ∆ v=a ∆t So ∆x = v0 ∆ t + ½ a ∆t ∆t ∆x = v0 ∆ t + ½ a ∆t2 v at 2 v0 1 t t0

  8. Example • An object falls from a height of 100 m and is of a shape where air friction can be ignored. • How long does it take to reach the ground? • What is its velocity just before impact? t =√ 2∆y/a = 4.5 s v = v0 + at = 0 - 9.8  4.5 = -44 m/s ∆y = v0t + 1/2at 2 a = -9.8m/s 2 v0= 0 ∆y = -100

  9. BUZZ Question A world’s land speed record was set by Colonel John P. Stapp when, on March 19, 1954, he rode a rocket-propelled sled that moved down a track at 1020 km/h. He and the sled were brought to a stop in 1.4 s. What acceleration did he experience? Express your answer in g units.

  10. Travelling down a track at 1020 km/h. Brought to a stop in 1.4 s. What acceleration did he experience? Express your answer in g units. Hint: use a=Δv/t=(vf -v0)/t a=[0-(1,020,000 m/60x60 s )]/1.4 s = -202.38 m/s2 in g units, means divide by 9.8 m/s2 Answer: about -21 g (decelerating)

  11. Horizontal axis g-force Untrained humans are able to tolerate 17 g “eyeballs-in” (acceleration) and about 12 g “eyeballs-out” (deceleration) for several minutes without loss of consciousness or apparent long-term harm. Stapp survived a peak “eyeballs-out” force of 46.2 g, and more than 25 g for 1.1 sec. Stapp lived another 45 years to age 89, but suffered lifelong damage to his vision.

  12. Another example (buzz) • An object is projected vertically upwards at a velocity of 20 m/s. (Ignore air friction) • How high does it get to? • How long before it falls back to ground? • What is its velocity just before impact? • What is its velocity and acceleration at the topmost point of its trajectory?

  13. Another example • An object is projected vertically upwards at a velocity of 20 m/s. (Ignore air friction) • How high does it get to? Consider first time to get to top: We have v0 = 20 m/s and vf= 0 m/s And we know a=-9.8 m/s2 Use vf= v0 + at, solve for t t= 20/9.8 = 2.04 s

  14. Another example • An object is projected vertically upwards at a velocity of 20 m/s. (Ignore air friction) • How high does it get to? We have t= 2.04 s Use y=y0 + v0 t + ½ a t2 y=0 + 20x 2.04 + ½ (-9.8) (2.04)2 = 40.8 -20.39 =20.4 m

  15. Another example • An object is projected vertically upwards at a velocity of 20 m/s. (Ignore air friction) • How long before it falls back to ground? Again use y=y0 + v0 t + ½ a t2 0=20.4 + 0 + ½ (-9.8) t2 t=(2x20.4/9.8 ) =2.04 s

  16. Another example • An object is projected vertically upwards at a velocity of 20 m/s. (Ignore air friction) • What is its velocity just before impact? • What is its velocity and acceleration at the topmost point of its trajectory? vf= v0 + at, = 0 + (-9.8) 2.04 = -20 m/s Velocity (top)= 0 m/s Acceleration = -9.8 m/s2

  17. Off the Line • So far we’ve developed the tools to model motion restricted to a line (“1-dimensional”) • The next step in being able to model realistic motion is to be restricted to a plane (“2-d..”) -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 y x

  18. Vectors • In 1-D motion displacement, velocity and acceleration all have one of two directions, represented by their arithmetic sign (+ or -) • In 2-D motion these quantities can have directions ranging over 360°, so simple signs won’t work. They will actually, but we have to get cunning.

  19. Vector Addition • Displacements can be added as vectors placed in a “tail-to-head” sequence. y The vector sum of a series of displacements is just the resultant displacement. d2 d1 d1 + d2 x

  20. Example (buzz) • A boat sails a course 3 km due North, then 3 km due East then 5 km 45° East of North. What is the sum of these displacements? y At the end of the second leg, x = 3000 m, y = 3000 m or distance r = √1.8x107m2 = 4240 m bearing  = 45° d3 N d2 d1 + d2 + d3 d1 x

  21. Example • A boat sails a course 3 km due North, then 3 km due East then 5 km 45° East of North. What is the sum of these displacements? y The third leg is a further 5000 m at bearing 45°, so the total displacement is 9240 m at= 45° d3 N d2 d1 + d2 + d3 d1 x

  22. Example • The third leg in this journey is equivalent to a sequence of two movements, one North (∆y) and the other East (∆x) [order unimportant]. ∆x = d sin ∆x = 5000 sin45° = 3536 m ∆y = 5000 cos45° = 3536 m ∆y = d cos d 

  23. Recap • We’ve done two things: • Added vectors to give a resultant • Worked out equivalent components of a vector • These operations can be represented: • graphically, using a scale diagram • algebraically, using components (more later) • These operations look pretty straightforward for displacement, but apply equally well to velocity and acceleration

  24. Question • If an object was thrown horizontally over a cliff, would you expect it to take longer to hit the ground below than if it was just dropped over?

  25. DEMMO

  26. Answer • The two objects take exactly the same time to fall. It is as if the vertical and horizontal components of the motion are independent. • Horizontal motion: acceleration = 0 (constant velocity) • Vertical motion: acceleration = -9.8 m/s2 (velocity increasing)

  27. Photographed at same time intervals

  28. Just before impact the yellow ball has a velocity v vx • vx is constant throughout ∆x = v0x t • vy increases with time ∆y = v0y t + 1/2at 2 v vy

  29. Example (buzz) x • If an object is thrown over a 100m cliff with an initially horizontal velocity of 20 m/s • how far out does it land? • what is its velocity at impact? y

  30. Example x • If an object is thrown over a 100m cliff with an initially horizontal velocity of 20 m/s • how far out does it land? • what is its velocity at impact? We know from before, that an object dropped from 100 m takes 4.5 s to land. In this time the horizontal distance travelled is ∆x = v0x t =20 m/s x 4.5 s = 90 m y

  31. Example

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