Acceleration Formulas

Acceleration Formula We have stated that our definition for acceleration is With this formula we are able to solve problems that deal with velocities and time. But what if a problem asks you to account for a distance? The above formula needs to be changed to accommodate this new variable.

Constant Acceleration Lets first start with a definition. Constant acceleration is non-changing acceleration (but the velocity is changing, so the object is either speeding up for slowing down). In other words it comes from a constant or linear slope of a velocity versus time graph. All the accelerations that we deal with are constant in form.

Calculating Velocity from Acceleration Lets start off by taking our original acceleration formula.

Calculating Velocity from Acceleration By rearranging our formula we can calculate the final velocity (vf ) if we know a, vo and t.

Calculating Velocity from Acceleration By rearranging our formula we can calculate the final velocity (vf ) if we know a, vf and t.

Calculating Velocity from Acceleration By rearranging our formula we can calculate the final velocity (vf ) if we know a, vo and t. Lose the Δ because time is given as a single number.

Calculating Velocity from Acceleration By rearranging our formula we can calculate the final velocity (vf ) if we know a, vo and t.

Calculating Velocity from Acceleration By rearranging our formula we can calculate the final velocity (vf ) if we know a, vo and t. FLIP

Calculating Velocity from Acceleration By rearranging our formula we can calculate the final velocity (vf ) if we know a, vo and t.

Calculating Velocity from Acceleration By rearranging our formula we can calculate the final velocity (vf ) if we know a, vo and t.

Calculating Velocity from Acceleration (part 2) If you are not given time but instead vf, vo, d or a, we need to use a different formula due to our original formula not having distance. Need to change the formula to incorporate d instead of t.

Calculating Velocity from Acceleration (part 2). Without going into the direct manipulation of the formula (see SNAP pages 51-53 if you are interested), we can arrange our formula to contain d.

Acceleration and Distance Many times you are asked to calculate the distance using acceleration formulae. Instead of going through the formula manipulation (which would explode your brain) lets just look at the formulae.

Acceleration and Distance If you are given vf, vo and t you can use the following:

Acceleration and Distance If you are given vo, a and t you can use the following:

Critical point By listing what you know and don’t know, you are able to pick the proper formula quickly and efficiently. Find the formula (it’s like a treasure hunt for physicists) that fits the variables that you have and the variable that you are trying to find

Problem #1 The Saturn rockets that carried astronauts to the Moon accelerated at a rate of 17.0m/s2, at take off. What is the velocity achieved when the rocket reaches an altitude of 1500m?

Problem #1 The Saturn rockets that carried astronauts to the Moon accelerated at a rate of 17.0m/s2, at tale off. What is the velocity achieved when the rocket reaches an altitude of 1500m? a =17.0m/s2 vo= 0m/s d = 1500m vf = ?

Problem #1 The Saturn rockets that carried astronauts to the Moon accelerated at a rate of 17.0m/s2, at tale off. What is the velocity achieved when the rocket reaches an altitude of 1500m? a =17.0m/s2 vo= 0m/s d = 1500m vf = ?

Problem #1 The Saturn rockets that carried astronauts to the Moon accelerated at a rate of 17.0m/s2, at tale off. What is the velocity achieved when the rocket reaches an altitude of 1500m? a =17.0m/s2 vo= 0m/s d = 1500m vf = ?

Problem #1 The Saturn rockets that carried astronauts to the Moon accelerated at a rate of 17.0m/s2, at tale off. What is the velocity achieved when the rocket reaches an altitude of 1500m? a =17.0m/s2 vo= 0m/s d = 1500m vf = ?

Problem #1 The Saturn rockets that carried astronauts to the Moon accelerated at a rate of 17.0m/s2, at tale off. What is the velocity achieved when the rocket reaches an altitude of 1500m? a =17.0m/s2 vo= 0m/s d = 1500m vf = ?

Problem #1 The Saturn rockets that carried astronauts to the Moon accelerated at a rate of 17.0m/s2, at tale off. What is the velocity achieved when the rocket reaches an altitude of 1500m? a =17.0m/s2 vo= 0m/s d = 1500m vf = ?

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s2. Calculate her stopping time.

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s. Calculate her stopping time. a = -3.40m/s2 vo= 30.0km/hr vf = 0 m/s t = ?

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s. Calculate her stopping time. a = -3.40m/s2 vo= 30.0km/hr vf = 0 m/s t = ? Need to convert km/hr into m/s

Problem #2 Janine is driving through a school zone at a speed of 30km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.4m/s. Calculate her stopping time. a = -3.40m/s2 vo= 30.0km/hr vf = 0 m/s t = ?

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s. Calculate her stopping time. a = -3.40m/s2 vo= 8.33m/s vf = 0 m/s t = ?

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s. Calculate her stopping time. a = -3.40m/s2 vo= 8.33m/s vf = 0 m/s t = ?

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s. Calculate her stopping time. a = -3.40m/s2 vo= 8.33m/s vf = 0 m/s t = ?

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s. Calculate her stopping time. a = -3.40m/s2 vo= 8.33m/s vf = 0 m/s t = ?

Problem #2 Janine is driving through a school zone at a speed of 30.0km/hr. She needs to stop suddenly to avoid a child’s soccer ball. Her brakes produce a deceleration of –3.40m/s. Calculate her stopping time. a = -3.40m/s2 vo= 8.33m/s vf = 0 m/s t = ?

Summary-Uniform Acceleration See p54 Remember v = d/t can NOT be used when there is acceleration Formula NOT on formula sheet: See example problems on p 54-55

Free Falling Objects (p 66) Objects accelerate uniformly (at the same rate) as they fall (add) near the surface of the earth acceleration due to gravity, ag = -9.80m/s2. Activity: drop a pencil and sheet of paper from the same height Repeat but this time crumple the paper into a tight ball

(p 67) acceleration of a freely falling object is called acceleration due to gravity Symbol: g or ag Near the earth’s surface this acceleration is approximately -9.80 m/s2. (add in margin) this number changes (decreases) as you move farther away from earth (into space) See Example Problems (p68): add the negative signs for the downward directions for “a” and “d” in questions 1, 2 and 3.

on Earth: ag = -9.80m/s2 down this means heavy and light objects fall at the same rate (if we ignore air friction)

Kinematics in 1D- vector quantity problems (p77) Displacement becomes zero when the ball returns to its starting point and will be down if it rolls beyond its starting point in the opposite direction (p77) direction along a line is indicated using + and – signs Right or up is + Left or down is – (also highlight this info on p 97) & see example problems 1-4