Presentation on theme: "One Dimensional Kinematics"— Presentation transcript:
1One Dimensional Kinematics Displacement (ŝ) - distance covered in a particular directionVelocity-- timed rate of change in displacement Avg. Vel = total displacement / total timev = ∆x∆t2.3,4,5Instantaneous Velocity: Velocity at any given instant!If the object is not accelerating then the avg. vel. = instant. vel.
23. 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: 2.3.3,42.4.6,7
3Acceleration-- timed rate of change in velocity Avg. Acceleration = total change in velocity / total timea = ∆v∆tInstantaneous Acceleration: the acceleration at any instant in time2.10,11,12*Notice Velocity and Acceleration vectors aren’t necessarily in the same direction!
411. 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
5Graphic Representation of Motion No motion (at rest):avxt2.6-9tt
6Motion at constant (non-zero) velocity: x2.6-9ttt
7Motion with constant (non-zero) acceleration: vx2.6-9ttt
12Instantaneous 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) = 0if the object is not accelerating (moving with constant velocity), then v(t) = avg. v = kGraph slope of curve– 2.13if the object is accelerating, then the instantaneous velocity equation can be found through the Limiting ProcessFrom now on: velocity mean instantaneous vel.!
13The Limiting Process-- Derivatives v = lim ∆x∆t ∆tv = dxdtTo find a derivative of a given equation:if x(t) = xnif x(t) = A (constant)2.24then dx = nxn-1dtthen dx / dt = 0the derivative of a constant term is 0 !
14The 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 = 1.0sx = 5(1.0) - 2(1.0)2 +4(1.0)3 = 7.0 mTo find velocity, take the first derivative of the displacement equation. Acceleration is the derivative of the velocity equation.
15B) v = dx / dt = 5 - 4t + 12t2 when t = 2.3 s v2.3 = 5 - 4(2.3) + 12(2.3)2 = 59.3 m/sC) a = dv / dt = dx / dt = t at t = 2.0 sa2.0 = (2.0) = 44 m/s2This 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!
17Equations for Motion with Constant Acceleration: VariablesEquationx vo v a tv = vo + at √ √ √ √x = xo +vot +.5at √ √ √ √v2 = vo2 + 2a(x - xo) √ √ √ √x = xo + .5(vo + v)t √ √ √ √x = xo + vt - .5at √ √ √ √
18Freefall: 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/s2When using this value in the previous equations, we will adopt the following conventions:2.26gravity will work along the y axis and up will be positiveconvention dictates that up is + and down is neg. so g is often used as -g in equations
19An 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 = 0vo2 = v2 - 2a∆yv = m/svo = 40.0 m/sa = -gvo = ?
20From the highest point: y = ?y - yo = (v2 - vo2) / 2ayo = 0vo = 0v = m/sy = ma = - gthe negative sign means that the object fell down 128 m from its highest point!