Acceleration

Changing Motion Objects with changing velocities cover different distances in equal time intervals.

Acceleration Rate at which an object’s velocity changes. –When the velocity changes at a constant rate, the object has constant acceleration. –Average Acceleration = change in velocity during a time interval –Instantaneous acceleration = acceleration at any instant.

Velocity-Time Graph A velocity-time graph can be used to find acceleration The slope of the velocity-time graph gives the acceleration. a =  v /  t a = (v f - v i ) / (t f - t i )

Positive and Negative Acceleration Positive acceleration means that the acceleration vector is in the positive direction Negative acceleration means that the acceleration vector is in the negative direction. –Positive and negative acceleration do not indicate speeding up or slowing down.

Ex. #1: A shuttle bus starts from rest and reaches a velocity of 9.00 m/s in 5.00 s. Find the average acceleration of the bus. Given: v i = 0 v f = 9.00 m/s  t = 5.00 s Find: a = ? a = (v f - v i ) /  t = (9.00 m/s - 0)/(5.00 s) = 1.80 m/s 2

Ex. #2: A motorcycle has an average acceleration of 15 m/s 2. How much time is required for it to reach 27 m/s from rest? Given: v i = 0 m/s v f = 27 m/s v f = 27 m/s a = 15 m/s 2 a = 15 m/s 2 Find:  t = ? a = (v f - v i ) /  t  t = (v f - v i ) / a = (27 m/s - 0 m/s) / (15 m/s 2 ) = (27 m/s - 0 m/s) / (15 m/s 2 ) = 1.8 s = 1.8 s

Velocity-Time Graph The area between the velocity-time curve and the x-axis gives the displacement of the object. –Add all areas to get total displacement. –Negative areas indicate negative displacement

Motion with Constant Acceleration Equations can be derived from a velocity-time graph velocity-time graph velocity-time graph 1. v f = v i + aΔt (slope of v-t graph) 2. d f = d i + v i Δt + ½ aΔt 2 (area under curve- triangle + rectangle) 3. Δd = ½ (v i + v f ) Δt (area under curve-trapezoid) 4. v f 2 = v i 2 + 2a Δd (combine 1 st and 2 nd equns to remove time)

Ex. 1: A boat on the Log Flume at Six Flags takes 2.0 s to slow down from 16.2 m/s to 1.7 m/s. What is the rate of acceleration of the boat? Given:  t = 2.0 s v i = 16.2 m/s v i = 16.2 m/s v f = 1.7 m/s v f = 1.7 m/s Find: a = ? v f = v i + a  t v f = v i + a  t (v f – v i ) /  t = a (1.7 m/s m/s) / 2.0 s = a = m/s 2

Ex. 2: How far does the same boat on the Log Flume travel while slowing down? Given:  t = 2.0 s v i = 16.2 m/s v i = 16.2 m/s v f = 1.7 m/s v f = 1.7 m/s Find:  d = ?  d = ½ (v i + v f )  t = ½ (16.2 m/s m/s)(2.0 s) = 18 m

Ex. 3: Back on the Log Flume, the next boat accelerates down the slide at 5.2 m/s 2. If the boat starts from rest, how long is the slide? Given: v i = 0 m/s v f = 16.2 m/s v f = 16.2 m/s a = 5.2 m/s 2 a = 5.2 m/s 2 Find:  d = ? v f 2 = v i 2 + 2a  d v f 2 = v i 2 + 2a  d (v f 2 – v i 2 ) / 2a =  d [(16.2 m/s) 2 - (0) 2 ] /(2*5.2 m/s 2 ) =  d = 25 m