Motion with Uniform Acceleration • When acceleration of a particle is
not
Equations of UARM •
If a particle has constant acceleration a, it follows that the average acceleration is also a.
•
The four basic UARM equations :
changing • Acceleration
is a constant non zero value
(UARM) • Acceleration
1. v=v0+at
is zero (UM)
2. Δx = ½ (v+v 0) * t 3. Δx = v0t + ½ at 2
2
2
4. v = v0 +2aΔx
One-dimensional Motion With Constant Acce Ac celer lerati ation on (U (UAM) AM)
v v v f o f o a t t t f o v
Uniform Accelerated Motion
Eq’n 1: where:
v f vo at vf = velocity of the body at a later time t v0 = initial velocity t = elapsed time
v f vo at
a = acceleration
Shows velocity as a function of acceleration and time
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Example 1 : VROOOOM! A motor bike passes a green traffic light while moving at a velocity of 6.00 m/s. It then accelerates at 0.300 m/s2 for 15.0 s. What is the bike’s velocity at 15.0s?
Uniform Accelerated Motion
Recall our equation for V av :
Manipulating this to get displacement:
Uniform Accelerated Motion
If the acceleration of a moving object is constant, its average velocity is:
One-dimensional Motion With Constant Acceleration (UAM) •
From Equation 2
v f vo at
v v f x vaveraget o t 2 1
x vot at
V0 is the initial velocity of the object while V is its final velocity Eq’n 2:
where:
Velocity changes uniformly
Represents the displacement x of the moving body at any instant of time t.
Uniform Accelerated Motion
Eq’n 3:
2
2
x vo t
1 2
2
at
v0 = initial velocity
Example 2 : Travelling Truck A truck, initially travelling at 8.33 m/s, accelerates at the constant rate of 1.50 m/s2. How far will the truck travel in 15.0s?
t = elapsed time a = acceleration ∆x = displacement
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Recall :
Example : Ready for take off!
•
Equation 1:
v f vo at
•
Equation 2:
x (
•
Equation 3 :
vo v 2
x vo t
1 2
• An
airplane accelerates from rest down a runway at 3.20 m/s 2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
)t
at 2
One-dimensional Motion With Constant Acceleration (UAM)
Uniform Accelerated Motion
Recall Equation 2: Eq’n 4:
vo v f t x vaveraget 2
And Equation 1:
v0 = initial velocity a = acceleration ∆x = displacement
v f vo a
v f vo 2a x 2
2
vo 2ax
where: vf = velocity of the body at a later time t
v f vo at t
2
v f
2
Represents the velocity of the moving body at any displacement
Summary of kinematic equations
Notes: Motion with Constant Acceleration •
If the velocity is constant (a=0), the displacement d is computed by multiplying by the velocity by the time Δx=v*t. x vo t
1 2
at 2
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Summary of kinematic equations EQUATION NUMBER
EQUATION
MISSING VARIABLE
1
v=v0+at
Δx
2
Δx = ½ (v+v 0) * t
a
3
Δx = v0t + ½ at 2
v
4
2
v =
v02+2aΔx
t
Example 3: Doggie on the Road •
Kat is driving around campus when she sees a dog in the middle of the road. She steps on the brakes and decelerates from a speed of 120 km/h to a complete stop during a displacement of 91 m at constant acceleration to avoid hitting the dog.
a. What is the acceleration? Given: v, v0, delta x Required: a Missing Quantity: t Equation to use?
b. How much time is required for the given decrease in speed? Given: v, v0, delta x, a Required: t Equation to use?
The Use of Missing Variables 1. Check for the given (supplied data) in the problem 2. Make sure it is stated that the motion is uniformly accelerated (or has uniform velocity) and is rectilinear (straight, narrow, flat paths) 3. Check for what is needed or required 4. Then check which variable is missing! •
After the abovementioned steps, determine the equation to use, then solve!
Example 4 : Road trip A car has an initial velocity of 20 m/s and an acceleration of -1.0 m/s2. (a) Find its displacement after the 10 s from the moment the acceleration begins. (b) How far will the car have gone when it comes to a stop?
Example 5: CREEPY WHITE LADY • On
a highway at night you see a white lady and brake your car to a stop with an acceleration of -5.6 m/s2. What is the car's stopping distance if its initial speed is a) 17 m/s? b) 28 m/s?
