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# Lecture 07

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### Lecture 07

1. 1. Linear AccelerationObjectives:• Introduce the concept of linear acceleration• Learn how to compute and estimate linear acceleration• Understand the difference between average and instantaneous acceleration• Learn to use the laws of constant acceleration Change in Velocity• Because velocity is a vector, a change in velocity is also a vector• Change in velocity has magnitude and direction vy (m/s) ∆v = change from vinitial to vfinal vintial vfinal vx (m/s) 1
2. 2. Computing a Change in Velocity• Compute change in velocity (∆v) by vector subtraction ∆v = vfinal – vinitial vy (m/s) vinitial vfinal vx (m/s) –vinitial –vinitial ∆v Linear Acceleration• The rate of change of linear velocity• Acceleration is a vector; has magnitude and direction change in velocity acceleration = change in time• Shorthand notation: v2 – v1 ∆v a = = t2 – t1 ∆t• Has units of length/time2 (e.g. m/s2, ft/s2) 2
3. 3. Computing Acceleration• direction of accel. = direction of change in velocity• magnitude of accel. = magnitude of change in velocity change in time• component of accel. = component of change in velocity change in time y ∆v y a = ∆v / ∆t / ∆t ∆v ∆v ∆vy ay = ∆vy / ∆t = θ a θ x x ∆vx ax = ∆vx / ∆t Acceleration as Change in Speed • Acceleration can reflect a change in the magnitude of velocity without a change in direction (e.g. a change in speed) vy ay v2 a = ∆v / ∆t ∆v v1 θ θ vx ax 3
4. 4. Acceleration as Change in Direction • Acceleration can reflect a change in the direction of velocity without a change in magnitude vy ay v1 ∆v v v2 ax v vx a = ∆v / ∆t Acceleration in General• In general, acceleration reflects a change in both the magnitude and direction of velocity• Can picture as accel. along original direction of motion and accel. away from original direction of motion vy ay a along dir. of motion a away from ∆v dir. of motion v1 v2 a = ∆v / ∆t vx ax 4
5. 5. Example Problem #1 A basketball player is moving forward at 6 m/s at the start of a jump. He leaves the ground 250 ms later with a forward velocity of 2.5 m/s and an upward velocity of 4 m/s What was his average acceleration between starting the jump and leaving the ground? Acceleration in 1-D• Velocity and acceleration in same direction: magnitude of velocity increases• Velocity and acceleration in opposite direction: magnitude of velocity decreases (deceleration) Velocity Acceleration Change in Velocity (+) (+) Increase in + dir. (+) (–) Decrease in + dir. (–) (–) Increase in – dir. (–) (+) Decrease in – dir. 5
6. 6. Example Problem #2 The instantaneous vertical velocity was determined at intervals of 100 ms during the movement shown. What was the average vertical acceleration over each 100 ms interval? 1.4 Time Velocity 1.2 (s) (m/s) vertical position (m) 1 0 0 0.8 0.1 4 0.6 0.2 6 0.4 0.3 3 0.2 0.4 -1 0 0 0.1 0.2 0.3 0.4 0.5 0.5 -5 time (s) Acceleration as a Slope• Graph x-component of velocity vs. time• x-component of acceleration from t1 to t2 = slope of the line from vx at t1 to vx at t2 vx (m/s) Slope : ∆vx / ∆t = a x ∆ vx ∆t t1 t2 time (s) 6
7. 7. Average vs. Instantaneous Accel.• Previous formulas give us the average acceleration between an initial time (t1) and a final time (t2)• Instantaneous acceleration is the acceleration at a single instant in time• Can estimate instantaneous acceleration using the central difference method: v (at t1 + ∆t) – v (at t1 – ∆t) a (at t1) = 2 ∆t where ∆t is a very small change in time Instantaneous Accel. as a Slope• Graph of x-component of velocity vs. time slope = instantaneous x-acceleration at t1 vx (m/s) slope = average x-acceleration from t1 to t2 ∆t t1 t2 time (s) 7
8. 8. Estimating Acceleration from Velocity vx (m/s) Identify points with zero slope = points with zero acceleration 0 Portions of the curve time (s) with positive slope have positive accel. (i.e. acceleration inax (m/s2) the + direction) Portions of the curve with negative slope 0 have negative accel. time (s) (i.e. acceleration in the – direction) Example Problem #3Below are velocity vs. time plots for two sprinters in the 100 m. Graph the acceleration of each sprinter.Who had the greater peak velocity? Acceleration? 14 12 10 Velocity (m/s) 8 6 4 JOHNSON 2 LEWIS 0 0 2 4 6 8 10 Time (s) 8
9. 9. Laws of Constant Acceleration• Under conditions where acceleration is constant: v2 = v 1 + a * ∆t d = v 1 * ∆ t + (½) a * ( ∆ t)2 v22 = v 12 + 2 a * d where: a = acceleration v 1 = velocity at initial (or first) time t1 v 2 = velocity at final (or second) time t2 d = displacement (change in position) between t1 and t2 ∆ t = change in time (= t2 – t1 ) Use + values for + direction, – values for – direction Special Cases• If acceleration (a) = 0: v 2 = v1 + a * ∆ t v 2 = v1 d = v1 * ∆t + (½) a * ( ∆t)2 d = v1 * ∆ t• If initial velocity (v1) = 0: v 2 = v1 + a * ∆ t v2 = a * ∆t d = v1 * ∆t + (½) a * ( ∆t)2 d = (½) a * ( ∆t)2 v22 = v12 + 2 a * d v22 = 2 a * d• If final velocity (v2) = 0: v 2 = v1 + a * ∆ t v1 = – a * ∆t v22 = v12 + 2 a * d v12 = – 2 a * d 9
10. 10. Example Problem #4A cyclist starts down a hill at 2 m/s and accelerates down the road at a constant rate of 0.6 m/s 2If the road down the hill is 200 m long, how fast is she moving at the bottom?If she then uses her brakes to decelerate at a constant 1.2 m/s 2, how much time will it take her to return to a velocity of 2 m/s?How far will she travel while she is decelerating? 10