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One Dimensional Kinematics

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Presentation on theme: "One Dimensional Kinematics"— Presentation transcript:

1 One Dimensional Kinematics
Displacement (ŝ) - distance covered in a particular direction Velocity-- timed rate of change in displacement  Avg. Vel = total displacement / total time v = ∆x ∆t 2.3,4,5 Instantaneous Velocity: Velocity at any given instant! If the object is not accelerating then the avg. vel. = instant. vel.

2 3. 2. A jogger is moving at a constant velocity of +3
3.2 A jogger is moving at a constant velocity of +3.0 m/s directly towards a traffic light that is 100 meters away. If the traffic light is at the origin, x = 0 m, what is her position after running 20 seconds? Homework: 2.3.3,4 2.4.6,7

3 Acceleration-- timed rate of change in velocity
 Avg. Acceleration = total change in velocity / total time a = ∆v ∆t Instantaneous Acceleration: the acceleration at any instant in time 2.10,11,12 *Notice Velocity and Acceleration vectors aren’t necessarily in the same direction!

4 11. 2. A baseball is moving at a speed of 40
A baseball is moving at a speed of m/s toward a baseball player, who swings his bat at it. The ball stays in contact with the bat for 5.00×10−4 seconds, then moves in essentially the opposite direction at a speed of 45.0 m/s. What is the magnitude of the ball's average acceleration over the time of contact? (These figures are good estimates for a professional baseball pitcher and batter.) Homework: ,7

5 Graphic Representation of Motion
No motion (at rest): a v x t 2.6-9 t t

6 Motion at constant (non-zero) velocity:
x 2.6-9 t t t

7 Motion with constant (non-zero) acceleration:
v x 2.6-9 t t t

8 v x a t t t Negative Acceleration! Decelerating! 2.6-9 negative slope!

9 v x t t Object is moving with a given positive velocity, slows to rest at a constant rate and continues to accelerate opposite to original direction! a 2.6-9 t

10 assume motion begins at x0 = 0 v x a t
2.6-9 t t

11 HW: Chapter 2: C.10, 11, 12,14 ,16 6.1, 2 7.1 8.1 9.1 9.2

12 Instantaneous Velocity and Acceleration
A mathematical formula [ v(t) ] can often be used for determining the velocity of an object at any given point during its motion. if the object is not moving, then v(t) = 0 if the object is not accelerating (moving with constant velocity), then v(t) = avg. v = k Graph slope of curve– 2.13 if the object is accelerating, then the instantaneous velocity equation can be found through the Limiting Process From now on: velocity mean instantaneous vel.!

13 The Limiting Process-- Derivatives
v = lim ∆x ∆t ∆t v = dx dt To find a derivative of a given equation: if x(t) = xn if x(t) = A (constant) 2.24 then dx = nxn-1 dt then dx / dt = 0 the derivative of a constant term is 0 !

14 The position of an object is given by the equation x = 5t - 2t2 + 4t3 where x is in meters and t is in seconds. A) Find the displacement (not the same as how far it traveled) of the object at t = 1.0 s. B) What the is velocity of the object when t = 2.3 s? C) What is the acceleration when t = 1.0 s? A) x = 5t - 2t2 + 4t3 when t = 1.0s x = 5(1.0) - 2(1.0)2 +4(1.0)3 = 7.0 m To find velocity, take the first derivative of the displacement equation. Acceleration is the derivative of the velocity equation.

15 B) v = dx / dt = 5 - 4t + 12t2 when t = 2.3 s
v2.3 = 5 - 4(2.3) + 12(2.3)2 = 59.3 m/s C) a = dv / dt = dx / dt = t at t = 2.0 s a2.0 = (2.0) = 44 m/s2 This object does NOT have a constant acceleration rate. The acceleration varies with time! When the acceleration is constant, simple Algebra can provide equations for x, v, a!

16 HW: Ch 2.13: 7, 9, 11, 14, 16

17 Equations for Motion with Constant Acceleration:
Variables Equation x vo v a t v = vo + at √ √ √ √ x = xo +vot +.5at √ √ √ √ v2 = vo2 + 2a(x - xo) √ √ √ √ x = xo + .5(vo + v)t √ √ √ √ x = xo + vt - .5at √ √ √ √

18 Freefall: only gravity affects the motion of the object.
The acceleration due to gravity varies from place to place on the earth, but for now we shall consider it to be a constant magnitude of: g = 9.8 m/s2 When using this value in the previous equations, we will adopt the following conventions: 2.26 gravity will work along the y axis and up will be positive convention dictates that up is + and down is neg. so g is often used as -g in equations

19 An object is held out over a cliff that is 50. 0 m high
An object is held out over a cliff that is 50.0 m high. The object is then thrown straight up and allowed to fall back down to the base of the cliff, where it hits with a speed of 50.0 m/s. What initial speed did was the object given and to what height above the base of the cliff did it rise? yo = 50.0 m (this means we have chosen the base of the cliff as our origin) y = 0 vo2 = v2 - 2a∆y v = m/s vo = 40.0 m/s a = -g vo = ?

20 From the highest point:
y = ? y - yo = (v2 - vo2) / 2a yo = 0 vo = 0 v = m/s y = m a = - g the negative sign means that the object fell down 128 m from its highest point!

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