NEWS

Motion in One Dimension Unit 1. Lesson 1 : Position, Velocity, and Speed Position : location of a particle with respect to a chosen reference point Displacement. - ppt download

By A Mystery Man Writer

Average Velocity : a particle’s displacement (  x) divided by the time interval (  t) during which that displacement occurs V = xx tt Average Speed : the total distance traveled divided by the total time interval required to travel that distance Avg speed = total distance total time Avg Velocity is a VECTOR QUANTITY Avg Speed is a SCALAR QUANTITY
Motion in One Dimension Unit 1
Lesson 1 : Position, Velocity, and Speed Position : location of a particle with respect to a chosen reference point Displacement : the change in position in some time interval  x = x f – x i Distance : the length of a path followed by a particle Displacement is a VECTOR QUANTITY Distance is a SCALAR QUANTITY
Example 1 Positiont(s)x(m) A030 B1052 C2038 D300 E40-37 F50-53 Find the displacement, average velocity, and average speed of the object between positions A and F.
Lesson 2 : Instantaneous Velocity and Speed Instantaneous Velocity : the limiting value of the ratio  x/  t as  t approaches zero v = xx tt lim  t  0 Instantaneous Speed : the magnitude of the instantaneous velocity
B) A race car starts from rest and speeds up to 100 m/s. C) A spacecraft drifts through space at constant velocity. Are there any points at which the instantaneous velocity has the same value as the average velocity over the entire motion . If so, identify the point(s)..
Answer : Slope of the tangent line drawn at the time in question. Example 2.
Lesson 3 : Acceleration Average Acceleration : the change in velocity (  v) divided by the time interval (  t) during which that change occurs a = vv tt = v f – v i t f – t i Instantaneous Acceleration : the limit of the average acceleration as  t appoaches zero a = vv tt lim  t  0
Answer : Slope of the tangent line drawn at the time in question..
A) Find the average acceleration in the time interval t =0 to t = 2.0 s. B) Determine the acceleration at t = 2.0 s..
Lesson 4 : Motion Diagrams v No acc Motion A v a Motion B v a Motion C
Graphs for each motion : x Motion A t v t a t x Motion B t v t a t x Motion C t v t a t
t graphs. A) v t x t a t B) v t x t a t v C) t x t a t.
t graphs. A) x t v t a t B) x t v t a t x C) t v t a t.
Answer : The area under the graph equals displacement Example 3.
Negative Area : Object is moving toward smaller x values and displacement is decreasing
Lesson 5 : Kinematic Equations a = vv tt = v f – v i t f – t i v f = v i + a  t v f = v i + at (for constant a) v = v i + v f 2 (for constant a)
 x = x f - x i  x = vt x f – x i = vt x f – x i = ½ (v i + v f )t x f = x i + ½ (v i + v f )t (for constant a)
x f = x i + ½ (v i + v f )t x f = x i + ½ [v i + (v i +at)]t v f = v i + at x f = x i + v i t + ½ at 2 (for constant a) x f = x i + ½ (v i + v f )t x f = x i + ½ (v i + v f ) ( v f - v i a ) v f 2 = v i 2 + 2a (x f – x i ) (for constant a)
Summary v f = v i + at (for constant a) v = v i + v f 2 (for constant a) x f = x i + ½ (v i + v f )t (for constant a) x f = x i + v i t + ½ at 2 (for constant a) v f 2 = v i 2 + 2a (x f – x i ) (for constant a)
A)What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop . B)If the plane touches down at position x i = 0, what is the final position of the plane .
One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.00 m/s 2. How long does it take her to overtake the car .
A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion..
Free-fall acceleration (g) = 9.80 m/s 2 +y-y a y = -g = m/s 2 For making quick estimates, use g = 10 m/s 2
The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. A)Determine the time at which the stone reaches its maximum height. B)Determine the maximum height..
E)Determine the velocity and position of the stone at t = 5.00 s. C)Determine the time at which the stone returns to the height from which it was thrown..
v = xx tt lim  t  0 = dx dt v = dx dt.
Find its velocity at t = 2 s..
a = vv tt lim  t  0 = dv dt a = dv dt.
Find its acceleration at t = 2 s. Acceleration is also the second derivative of position. a = vv tt lim  t  0 = dv dt = d2xd2x dt 2.
Find its acceleration at t = 3 s..
The Integral or Antiderivative v t titi tftf tntn vnvn  x n = v n  t n = shaded area  x =  v n  t n = total area
As rectangles get narrower,  t n  0  x = lim  t n  0  v n  t n Displacement = area under v-t graph This is called the Definite Integral lim  t n  0  v n  t n = v(t) dt  titi tftf
Integrating Velocity v(t) = dx dt dx = v dt  xixi xfxf dx =  titi tftf v dtx f – x i =  titi tftf v dt
Integrating Acceleration a(t) = dv dt dv = a dt  vivi vfvf dv =  titi tftf a dtv f – v i =  titi tftf a dt
Finding Antiderivatives If k is a constant,  k dx = kx + C  x n dx = x n+1 n C (n = 1)
If the object starts from rest, find its position at t = 5s..
Find the position and velocity at arbitrary times..