AccelerationAcceleration

Changing MotionChanging Motion

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

AccelerationAcceleration

Rate at which an object’s velocity Rate at which an object’s velocity changes.changes.– When the velocity changes at a constant rate, When the velocity changes at a constant rate,

the object has constant acceleration.the object has constant acceleration.– Average Acceleration = change in velocity Average Acceleration = change in velocity

during a time intervalduring a time interval– Instantaneous acceleration = acceleration at Instantaneous acceleration = acceleration at

any instant.any instant.

Velocity-Time GraphVelocity-Time Graph

A A velocity-time graph velocity-time graph can be used to find can be used to find accelerationacceleration

The slope of the The slope of the velocity-time graph velocity-time graph gives the gives the acceleration.acceleration.

aa = = v / v / tt

a = (va = (vf f - v- vii) / ) / (t(tff - t - tii ) )

Positive and Negative AccelerationPositive and Negative Acceleration

Positive acceleration Positive acceleration means that the means that the acceleration vector is in acceleration vector is in the positive directionthe positive direction

Negative acceleration Negative acceleration means that the means that the acceleration vector is in acceleration vector is in the negative direction.the negative direction.– Positive and negative Positive and negative

acceleration acceleration do notdo not indicate speeding up or indicate speeding up or slowing down.slowing down.

Ex. #1:Ex. #1: A shuttle bus starts from rest A shuttle bus starts from rest and reaches a velocity of 9.00 m/s in and reaches a velocity of 9.00 m/s in 5.00 s. Find the average acceleration of 5.00 s. Find the average acceleration of the bus.the bus.

Given:Given: vvii = 0 = 0

vvff = 9.00 m/s = 9.00 m/s

tt = 5.00 s = 5.00 s

Find:Find: aa = ? = ?

a = (va = (vff - v - vii)) / / tt

= (9.00 m/s - 0)/(5.00 s) = (9.00 m/s - 0)/(5.00 s)

= 1.80 m/s= 1.80 m/s22

Ex. #2:Ex. #2: A motorcycle has an average A motorcycle has an average acceleration of 15 m/sacceleration of 15 m/s22. How much time . How much time is required for it to reach 27 m/s from is required for it to reach 27 m/s from rest?rest?

Given:Given: vvii = 0 m/s = 0 m/s

vvff = 27 m/s = 27 m/s

aa = 15 m/s = 15 m/s2 2

Find:Find: tt = ? = ?

a = (va = (vff - v - vii)) / / tt

t = (vt = (vff - v - vii) / a) / a

= (27 m/s - 0 m/s) / (15 m/s= (27 m/s - 0 m/s) / (15 m/s22))

= 1.8 s= 1.8 s

Velocity-Time GraphVelocity-Time Graph

The The areaarea between the between the velocity-time curve velocity-time curve and the x-axis gives and the x-axis gives the the displacementdisplacement of of the object.the object.– Add all areas to get Add all areas to get

total displacement.total displacement.– Negative areas Negative areas

indicate negative indicate negative displacementdisplacement

Motion with Constant AccelerationMotion with Constant Acceleration

Equations can be derived from a Equations can be derived from a velocity-time graphvelocity-time graph

1.1. vvff = v = vii + a + aΔΔtt (slope of v-t graph)(slope of v-t graph)

2.2. ddff = d = di i + + vvi i ΔΔt + ½ t + ½ aaΔΔtt22 (area under curve-(area under curve-

triangle + rectangle)triangle + rectangle)

3.3. ΔΔd = ½ (d = ½ (vvi i + v+ vff) ) ΔΔtt (area under curve-trapezoid)(area under curve-trapezoid)

4.4. vvff22 = v = vii 22 + 2a + 2a ΔΔd d (combine 1(combine 1stst and 2 and 2ndnd equns to equns to

remove time)remove time)

Ex. 1:Ex. 1: A boat on the A boat on the Log FlumeLog Flume at Six Flags at Six Flags takes 2.0 s to slow down from 16.2 m/s to takes 2.0 s to slow down from 16.2 m/s to 1.7 m/s. What is the rate of acceleration of 1.7 m/s. What is the rate of acceleration of the boat?the boat?

Given: Given: tt = 2.0 s = 2.0 s

vvii = 16.2 m/s = 16.2 m/s

vvff = 1.7 m/s= 1.7 m/s

Find: Find: aa = ? = ?

vvff= v= vii + a + a tt

(v(vff – v – vii) / ) / t = at = a

(1.7 m/s -16.2 m/s) / 2.0 s = (1.7 m/s -16.2 m/s) / 2.0 s = aa

= - 7.3 m/s= - 7.3 m/s22

Ex. 2:Ex. 2: How far does the same boat on the How far does the same boat on the Log FlumeLog Flume travel while slowing down? travel while slowing down?

Given: Given: t = 2.0 s t = 2.0 s

vvii = 16.2 m/s = 16.2 m/s

vvff = 1.7 m/s = 1.7 m/s

Find: Find: dd = ? = ?

d = ½ (vd = ½ (vii + v + vff) ) tt

= ½ (16.2 m/s + 1.7 m/s)(2.0 s)= ½ (16.2 m/s + 1.7 m/s)(2.0 s)

= 18 m= 18 m

Ex. 3:Ex. 3: Back on the Back on the Log FlumeLog Flume, the next boat , the next boat accelerates down the slide at 5.2 m/saccelerates down the slide at 5.2 m/s22. If the . If the boat starts from rest, how long is the slide?boat starts from rest, how long is the slide?

Given: Given: vvii = 0 m/s = 0 m/s

vvff = 16.2 m/s = 16.2 m/s

aa = 5.2 m/s = 5.2 m/s22

Find: Find: dd = ? = ?

vvff 22 = v = vii 22 + 2a + 2add

(v(vff 22 – v – vii 22) / 2a = ) / 2a = dd

[(16.2 m/s)[(16.2 m/s)22 - (0) - (0)22] /(2*5.2 m/s] /(2*5.2 m/s22) = ) = dd

= 25 m= 25 m