a = 0 m/s 2 Motion with Constant Velocity x f = x 0 + vt Motion with Constant Acceleration v f = v 0...

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Transcript of a = 0 m/s 2 Motion with Constant Velocity x f = x 0 + vt Motion with Constant Acceleration v f = v 0...

a = 0 m/s2

Motion with Constant Velocityxf = x0 + vt

Motion with Constant Accelerationvf = v0 + at

xf = x0 + v0t + ½at2

vf2 = v0

2 + 2aΔx

Equations of Motion!

#2 can also be written as Δx = v0t + ½at2, since Δx = xf – x0

ttThe “big three”

AP Level Kinematics Problem

A rock is dropped off a cliff and falls the first half of the distance to the ground in 3.0 seconds. How long will it take to fall the second half? Ignore air resistance.

(Hint: Figure out the height of the cliff first!)

Choose an Origin and Positive Direction

Easiest choice!

y0 = 0 ma = 9.8 m/s2

Δy will be positive

How High is the Cliff?

A rock is dropped off a cliff and falls the first half of the distance to the ground in 3.0 seconds.

For the first half of the fall, we can use Δy = v0t + ½at2

where Δy is half the height of the cliff and t = 3 s

This gives Δy = ½ * (9.8 m/s2) * (9 s2) = 44.1 m

Therefore, the cliff is 88.2 m high.

Initial: Top of cliff (v0 = 0) Final: Halfway down

How long does it take to fall the second half?

Now, we need to use Δy = v0t + ½at2, but with different initial and final conditions.

Initial: Halfway down (the rock will have a nonzero v0)Final: Just about to hit the ground.

Δy = 44.1 m (second half is just as far as first half)To find the rock’s velocity halfway down, we must usevf = v0 + at, and we get vhalfway = 9.8 m/s2 * 3s = 29.4 m/s

Finishing the job

Δy = 44.1 mv0 = 29.4 m/s

a = 9.8 m/s2

Δy = v0t + ½at2

44.1 = (29.4)t + 4.9 t2

Sometimes (semi-rarely), you will need to use the quadratic equation to calculate time when the object has an initial position, initial velocity and a nonzero acceleration.

One of your homework problems uses it; make sure you also know the quadratic equation by heart.

Remember to check if a, b or c are negative.

44.1 = (29.4)t + 4.9 t2

Rearrange to get all terms on one side4.9 t2 + 29.4 t – 44.1 = 0

a = 4.9 b =29.4 c = -44.1

Using quadratic, we gett = -7.24 s or 1.24 s

The first half takes 3 seconds…The second half takes only 1.24 seconds!

Why?

But WAIT!!!!!!

• There’s an easier, non-quadratic-ier way to solve it!

• Once you find that the cliff is 88.2 m tall, and you know that the first half of the drop takes 3 seconds…

• Find the total time of the whole drop, and subtract 3 seconds!

• This should get you the same result – try it!

Tips for Solving Kinematics Problems

• Read critically (“dropped”, “comes to stop”, “highest point reached”)

• Direction of a vector determines its sign.• Record all of the given information• Memorize the big three!• You will not have a calculator on the AP Exam

multiple choice, so if the calculations become too hard, go back and try a different approach

Final TipIf you encounter a toughie, try to step back,

breathe and look at the scenario as a whole. There may be a shortcut or technique that makes the problem easier and saves you time!

This can’t really be put into a series of steps, you just need to practice by studying hard and spending time on your homework

Meet the Position vs Time Graph

Interpreting x vs t graphs

The slope of the tangent line represents the instantaneous velocity of the object at anyone instant in time

The slope of a secant line represents the average velocity of the object between two moments in time.

vav = Δx/Δt

Example from 2008 Exam

The figure to the left shows the position vs. time graphs for two objects, A and B, that are moving along the same axis. The two objects have the same velocity

(A) at t = 0 s.(B) at t = 2 s.(C) at t = 3 s.(D) at t = 4 s.(E) never

Example from 2008 Exam

The figure to the left shows the position vs. time graphs for two objects, A and B, that are moving along the same axis. The two objects have the same velocity

(A) at t = 0 s.(B) at t = 2 s.(C) at t = 3 s.(D) at t = 4 s.(E) never

vv

Example from 2009 Exam

The graph below represents position vs. time for a sprinter at the start of a race. Her average speed during the interval between 1 second and 2 seconds is most nearly:

(A) 2 m/s(B) 4 m/s(C) 5 m/s(D) 6 m/s(E) 8 m/s

Example from 2009 Exam

The graph below represents position vs. time for a sprinter at the start of a race. Her average speed during the interval between 1 second and 2 seconds is most nearly:

(A) 2 m/s(B) 4 m/s(C) 5 m/s(D) 6 m/s(E) 8 m/s

However, x vs t graphs also have a SECRET PROPERTY hidden within

them...

…but I’m not going to tell you until tomorrow.

JUSTKIDDINGHEREWEGO• The concavity of the x vs t graph tells you the

acceleration of the object!