Acceleration
What is it?
Acceleration – Change in velocity divided by the time interval in which the change occurred.
Any change in velocity results in an acceleration. THIS INCLUDES JUST A CHANGE IN DIRECTION
Graph of Acceleration
• Slope of graph is the acceleration
• Constant Acceleration – the acceleration of a body does not change.
• Average acceleration ­ the acceleration of a body over a time interval.
• Instantaneous acceleration ­ value at that instant.
• Will work with constant acceleration only
Acceleration = (final velocity – initial velocity)
time
a = (vf – vi) t
Units of m/s/s or ft/s/s
Acceleration example
A car goes from 10 m/s to 20 m/s in 20 s. Find the average acceleration of the car.
Given: vi = vf = Equation : a = (vf – vi) t
Solve : a = t = Practice 2 Steve is moving along the sidewalk at 0.5 m/s. He sees a quick meal ahead, lunging at a mouse and accelerating at 1.5 m/s2. Steve reaches a velocity of 5 m/s when he catches the mouse. How long did it take Steve to reach the mouse? Given: vi = vf = a = Equation : a = (vf – vi) t
t=
Solve : t=
Jerry is racing his slot car and needs to avoid a sudden pile of cars that has appeared in his path. His car is traveling at 16 m/s and is able to slow with an acceleration of ­32 m/s/s. Jerry only has 0.4 s to stop his car. Determine how fast the car is traveling at the end of the 0.4s.
Given: v = i
t = Equation : a = (vf – vi) t
vf = Solve : vf = a = acceleration is constant
acting down the incline
Using Acceleration
• For objects undergoing uniform acceleration, the velocity will increase in linear amounts each second but the position of the object won’t. • Position increases as a quadratic amount • This happens due to the constant change in velocity.
The acceleration on the graph is constant. It is uniform motion.
v = x/t
a = (vf ­ vi) / t
• Equations to use only if acceleration is constant
vf=vi + at
x= ½ ( vi + vf)t
x= vit + ½ at2 displacement when the initial
and final velocity are for the
exact time given
Example Problem 1
Ima Hurryin is approaching a stoplight moving with a velocity of 30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop.
• If Ima's acceleration is ­8.00 m/s2 , how long does it take Ima to stop?
• Determine the displacement of the car during the skidding process.
• How long does it take Ima to stop?
Given : a = vi =
Equation: Solve :
vf =
t = ?
a = (vf – vi) / t
t = (vf – vi) / a
t = Determine the displacement of the car during the skidding process.
Given : a = vi =
vf =
t = Equation: x = vit + ½ a t2
Solve : x = x = ?
Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for 4.10 seconds.
• Determine the displacement of Ben's car during this time period.
• How fast is Ben going after this time?
Determine the displacement of Ben's car during this time period.
Given :
t = a = x = ?
Equation: x = vit + ½ a t2
Solve: x= vi = How fast is Ben going after this time?
Given : t = vi = a = vf = ?
Equation: a = (vf – vi) / t
Solve: v = f
vf = vi + at