Acceleration Science 1206

Acceleration: Symbol a, units m/s2 or km/h/s A measurement of the rate of CHANGE IN SPEED per unit time. NOTE: A “change in speed” could mean INCREASING or DECREASING speed. LINK

a = v2 – v1 t a = acceleration, in m/s2 or km/h/s There are TWO equations used for acceleration: Equation 1: Where a = acceleration, in m/s2 or km/h/s V2 = final velocity, in m/s or km/h V1= initial velocity, in m/s or km/h t = time, in s or h USE WHEN SOLVING FOR a or t V a = v2 – v1 t a t

1 m/s2 = speed changes by 1 m/s for every second of passing time NOTE: 1 m/s2 = speed changes by 1 m/s for every second of passing time ALSO NOTE: Negative (-) acceleration could mean one of two possibilities – (1) an object is slowing down OR (2) an object is going in the opposite direction. LINK t v 1 s 1 m/s 2 s 2 m/s 3 s 3 m/s

SAMPLE ACCELERATIONS - Just for fun  Acceleration in m/s2 Event 0.5 Elevator 1.6 Freefall acceleration on the moon 9.8 Freefall acceleration on Earth 3.7 Freefall acceleration on Mars 24.8 Freefall acceleration on Jupiter 270 Freefall acceleration on the Sun 10-40 Manned rocket at launch 20 Space shuttle 20-50 Roller coaster 80 Limit of sustained human tolerance 600 Airbag deployment 106 Bullet in the barrel of a gun

Example 1: Jacob wakes up late and realizes he is missing his Science 1206 class! Mortified, he rushes to his car and goes from being parked and at rest to 60.0 km/h in 2.00 s. What is his acceleration? V2= 60.0 km/h V1 = 0 km/h t = 2.00 s a = ?

Example 2: Acceleration due to gravity is 9.80 m/s2. If a ball is dropped off a building, what will be its change in speed after 10.0 s? a = 9.80 m/s2 t = 10.0 s v = ?

Example 3: Alex goes down Marble on her snowboard changing speed from 2.5 m/s to 4.4 m/s. If her average acceleration is 0.40 m/s2, how long is she on the hill? V2= 4.4 m/s V1 = 2.5 m/s t = ? a = 0.40 m/s2

USE WHEN SOLVING FOR v2 or v1 Equation 2 – This equation is derived from the first acceleration equation. v2 = v1 + at USE WHEN SOLVING FOR v2 or v1

Example 4: Scott is coming down the hallway at an initial speed of 0.50 m/s when he spots Mr. Barrett and accelerates at a rate of 0.33 m/s2 for 9.5 s. What is his final speed? V2= ? V1 = 0.50 m/s t = 9.5 s a = 0.33 m/s2

Example 5: A snowmobile at the uphill drag races reaches a final speed of 22.5 m/s after accelerating at 1.2 m/s2 for 17 s. What was the initial speed of the snowmobile? V2= 22.5 m/s V1 = ? t = 17 s a = 1.2 m/s2

SUMMARY: a = v2 – v1 t = v2 – v1 t a v2 = v1 + at v1 = v2 - at Use Equation 1 when you are solving for a or t a = v2 – v1 t = v2 – v1 t a Use Equation 2 when you are solving for v1 or v2 v2 = v1 + at v1 = v2 - at

Example 6: Neil is driving on the highway at a speed of 100 km/h when he sees a mink crossing the road and slams on the brakes and screeches to a stop just missing the mink! If the total elapsed time is 5.0 s, what is his average acceleration? V2= 0 km/h V1 = 100 km/h t = 5.0 s a = ?

HOMEWORK!!!! Complete WORKSHEETS 15 and 16 on p. 34 and 35 of your booklet.

Speed-Time Graphs y-axis – speed x-axis – time   The slope of a speed-time graph is equal to the average acceleration. Slope = a = y2-y1 = rise x2-x1 run

Describing Motion Using Speed-Time Graphs (What does the line mean?) Increasing speed Constant acceleration Constant speed No acceleration  

Decreasing speed Constant acceleration

Area Under the Line The area under the line of a speed-time graph is equal to the DISTANCE TRAVELLED. Area Under Line = Distance Area of a rectangle = l x w = 30 m/s x 20 s = 600 m

Examples 1) Tyrone is cruising down the Humber River in his canoe at a constant speed for 1.5 h. If the speed was 5.0 km/h, draw a speed-time graph of this situation, and calculate the distance that was travelled in this time.

2) Alaina speeds by in her speedboat, toppling over Tyrone’s canoe with the waves, and she accelerates away for 20 s. How far away did she get in that time? 0.0 5.0 10.0 15.0 20.0 25.0 10 Time (s) Speed (m/s)

Homework Please complete Worksheets 17 and 18 on p. 39 and 40 of your booklet!

aA = slope = 3.0 m/s2 aB = slope = 1.25 m/s2 dA = area under the line = area  = bh 2 = 150 m dB = 62.5 m

a = slope = 0 m/s2 d (0-8 s) = area under the line = area  = l x w = 8 x 18 = 144 m d (0-16 s) = 16 x 18 = 288 m

a = slope = 1.25 m/s2 d (0-8 s) = area under the line = area  + area  = 40 m + 80 m = 120 m d (0-16 s) = 1600 m + 160 m = 320 m

InTERPRETING GRAPHS

In which of the following graphs below are both runners moving along at the same speed? Explain your answer.

The distance- time graphs below represent the motion of a car The distance- time graphs below represent the motion of a car. Match the descriptions with the graphs. Explain your answers.

The speed-time graphs below represent motion of a car The speed-time graphs below represent motion of a car. Match the descriptions with the graphs. Explain your answers.

Please complete WORKSHEETS 19 and 20 on pg. 45 & 46 of your booklet!!! HOMEWORK!!! Please complete WORKSHEETS 19 and 20 on pg. 45 & 46 of your booklet!!!