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Equations of Uniform Accelerated Motion

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Equations of Uniform Accelerated Motion

AP Physics CMrs. Coyle

- Motion with constant acceleration
- Straight line
- Same direction

- Velocity v= vo+ at
- Position x= xo + vot + ½ at2
- v2 = vo2 + 2a(x-xo)
Remember:

Displacement= Dx = x-xo

- vavg= ½ (vo+v)
- Dx = ½ (vo+ v)t
- Assume that ti=0

Position (m)

x= xo + vot + ½ at2

Parabola

o

Time (s)

Slope of Tangent at a given time= Instantaneous Velocity at that time

Example of Velocity vs Time (Positive Acceleration)

v= vo+ at

Velocity (m/s)

o

Time (s)

Slope of Line= Acceleration

Area Under Line=Displacement

Example of Acceleration vs Time (Const. a)

Acceleration (m/s2)

o

Time (s)

Area under line = Change in Velocity

How do we derive Dx = ½ (vo+ v)t

from the graph?

v

Velocity (m/s)

vo

o

Time (s)

t

Hint: Area Under the Line=Displacement Δx

Position (m)

Parabola

o

Time (s)

Example of Velocity vs Time (Negative Acceleration)

Velocity (m/s)

o

Time (s)

Slope of Line= Acceleration

Area Under Line=Displacement

Example of Acceleration vs Time (Negative a)

Acceleration (m/s2)

o

Time (s)

Area under line = Change in Velocity

Hint: Start with Dx = ½ (vo+ v)t

and then substitute for vthat v= vo+at.

- Hint: Start with Dx = ½ (vo+ v)t
then substitute for t= (v– vo) /a

A ball initially stationary, accelerates at 0.25m/s2 down a 2m inclined plane. It then rolls up another incline, where it comes to rest after rolling up 1m.

a) What is the speed of the ball at the bottom of the incline and how much time did this take?

b) What is the acceleration along the second plane?

Answer: a) 1m/s, 4sec, b) -0.5m/s2

A Mustang travelling with a constant velocity of 35m/s, passes a stationary police car.

The reaction time of the officer was 2.5sec and he then accelerates at 5.0 m/s2 to catch the Mustang. How long does it take for the police car to catch the Mustang?

Answer: 15.8sec