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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 Instantaneous Velocity: Velocity at any given instant! If the object is not accelerating then the avg. vel. = instant. vel.

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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, ,7

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

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11.2 A baseball is moving at a speed of 40.0 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

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Graphic Representation of Motion No motion (at rest): x t v t a t

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Motion at constant (non-zero) velocity: x t v t a t

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Motion with constant (non-zero) acceleration: x t v a t t

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x t Decelerating! v t negative slope! a t Negative Acceleration!

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

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x t v t a t v t x t assume motion begins at x 0 = 0 a t

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HW: Chapter 2: C.10, 11, 12,14,16 6.1,

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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 if the object is accelerating, then the instantaneous velocity equation can be found through the Limiting Process From now on: velocity mean instantaneous vel.!

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

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The position of an object is given by the equation x = 5t - 2t 2 + 4t 3 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 - 2t 2 + 4t 3 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.

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B) v = dx / dt = 5 - 4t + 12t 2 when t = 2.3 s v 2.3 = 5 - 4(2.3) + 12(2.3) 2 = 59.3 m/s C) a = dv / dt = dx / dt = t at t = 2.0 s a 2.0 = (2.0) = 44 m/s 2 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!

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

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Equations for Motion with Constant Acceleration: Equation Variables x v o v a t v = v o + at √ √ √ √ x = x o +v o t +.5at 2 √ √ √ √ v 2 = v o 2 + 2a(x - x o ) √ √ √ √ x = x o +.5(v o + v)t √ √ √ √ x = x o + vt -.5at 2 √ √ √ √

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Freefall 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/s 2 When using this value in the previous equations, we will adopt the following conventions: å gravity will work along the y axis and up will be positive å convention dictates that up is + and down is neg. so g is often used as -g in equations

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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? y o = 50.0 m (this means we have chosen the base of the cliff as our origin) v = m/s a = -g v o = ? v o 2 = v 2 - 2a∆y y = 0 v o = 40.0 m/s

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

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