Motion with Constant Acceleration
Constant Acceleration
• In many practical situations:
– The magnitude of the acceleration is
uniform (constant)
– The motion is in a straight line
• It’s useful to derive some equations
which apply in this special case
ONLY!!!
– The kinematic equations for constant
(uniform) acceleration in one dimension.
Constant Acceleration
• Derivation is in the text. Also partially on the next
slide! Read on your own!
• In the derivation, its useful to change notation slightly.
• Note: My preferred notation is slightly different than our text!!
t1  0 = time when the problem begins
x1  x0 = initial position (at t1 = 0, often x0 = 0)
v1  v0 = initial velocity (at t1 = 0)
t2  t = time when we wish to know other quantities
x2  x = position at time t
v2  v = velocity at time t
a  acceleration = constant
(average & instantaneous accelerations are equal)
• By definition we have:
–Average velocity:
v = (x - x0)/t
(1)
–Acceleration (average = instantaneous):
a = (v - v0)/t
(2)
–Average velocity (another form):
v = (½)(v + v0)
(3)
Constant Acceleration Equations
Note Again: My preferred notation is
slightly different than in our text!!
• Results (1-dimensional motion only!):
v = v0 + at
(1)
x = x0 + v0 t + (½)a t2 (2)
2
2
v = (v0) + 2a (x - x0) (3)
v = (½) (v + v0)
(4)
NOT VALID UNLESS a = CONSTANT!!!
Often, x0 = 0. Sometimes v0 = 0
All we need for 1 dimensional constant-acceleration problems:
NOT VALID UNLESS a = CONSTANT!!!
Physics and Equations
IMPORTANT!!!
• Even though these equations & their applications
are important, Physics is not a collection of
formulas to memorize & blindly apply!
• Physics is a set of PHYSICAL PRINCIPLES.
• Blindly searching for the “equation which will
work for this problem” can be DANGEROUS!!!!
• On exams, you get to have an 8.5´´  11´´ sheet
with anything written on it (both sides) you wish.
On quizzes, I will give you relevant formulas.
Problem Solving Strategies
1. Read the whole problem. Make sure you understand it.
2. Decide on the objects of study & what the time interval is.
3. Sketch a diagram & choose coordinate axes.
4. Write down the known quantities, & the unknown ones needed.
5. What physics applies? Plan an approach to a solution.
6. Which equations relate known & unknown quantities?
Are they valid in this situation? Solve algebraically for
the unknown quantities, & check that your result is
sensible (correct dimensions).
7. Calculate the solution, round it to appropriate number
of significant figures.
8. Look at the result - is it reasonable? Does it agree with
a rough estimate?
9. Check the units again.
Bottom Line:
THINK!
DO NOT BLINDLY
APPLY FORMULAS!!!!
Example: Runway Design
You’re designing an airport. A plane that will use this airport must reach a
speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If
the runway is x = 150 m long, can this plane reach the speed of before it
runs off the end of the runway? (b) If not, what is the minimum length
required for the runway?
Example: Runway Design
You’re designing an airport. A plane that will use this airport must reach a
speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If
the runway is x = 150 m long, can this plane reach the speed of before it
runs off the end of the runway? (b) If not, what is the minimum length
required for the runway?
Table of Knowns & Unknowns
Example: Runway Design
You’re designing an airport. A plane that will use this airport must reach a
speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If
the runway is x = 150 m long, can this plane reach the speed of before it
runs off the end of the runway? (b) If not, what is the minimum length
required for the runway?
Table of Knowns & Unknowns
Solutions
(a) v2 = (v0)2 + 2a(x – x0)
v2 = 0 + 2(2.0)(150 – 0) = 600 m/s2
So v = (600)½ = 24.5 m/s
Note that this means take the
square root! That obviously matters!
Example: Runway Design
You’re designing an airport. A plane that will use this airport must reach a
speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If
the runway is x = 150 m long, can this plane reach the speed of before it
runs off the end of the runway? (b) If not, what is the minimum length
required for the runway?
Table of Knowns & Unknowns
Solutions
(a) v2 = (v0)2 + 2a(x – x0)
v2 = 0 + 2(2.0)(150 – 0) = 600 m/s2
So v = (600)½ = 24.5 m/s
Note that this means take the
square root! That obviously matters!
(b) Use Eq. (3) again with
v = vmin = 27.8 m/s. Solve for
x – x0 = [v2 – (v0)2]/(2a)
x = [(27.8)2 – 0]/[2(2.0)]
So x = 193 m.
To be safe, make the runway 200 m
long!
Example: Acceleration of a Car
How long does it take a car to cross a 30 m wide intersection
after the light turns green if it accelerates at a constant 2.0 m/s2?
Obviously, it
starts from rest!!
Known: x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2
Wanted: t.
Example: Acceleration of a Car
How long does it take a car to cross a 30 m wide intersection
after the light turns green if it accelerates at a constant 2.0 m/s2?
Obviously, it
starts from rest!!
Known: x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2
Wanted: t. Use: x = x0 + v0t + (½)at2 = 0 + 0 + (½)at2

t = (2x/a)½ = 5.48 s
NOTE! The square root obviously matters!
Example: Air Bags
You need to design an air bag system that can protect the driver at a
speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the
driver. How does the use of a seat belt help the driver?
Known: x0 = v0 = 28 m/s
v=0
Car obviously stops when crash ends! 
Wanted unknown: t.
But we don’t know acceleration a or
distance x either!
(1)
(2)
(3)
(4)
Example: Air Bags
You need to design an air bag system that can protect the driver at a
speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the
driver. How does the use of a seat belt help the driver?
Known: x0 = v0 = 28 m/s
v=0
Car obviously stops when crash ends! 
Wanted unknown: t.
But we don’t know acceleration a or
distance x either! Estimate x = 1.0 m
This has to be a 2 step problem! First, use
(2) to solve for a: 0 = (v0)2 + 2a(x – 0) so
a = - (v0)2∕(2x) = - (28)2 ∕(2) = - 390 m/s2
(1)
This is a HUGE acceleration!!
(2)
(3)
(4)
Example: Air Bags
You need to design an air bag system that can protect the driver at a
speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the
driver. How does the use of a seat belt help the driver?
Known: x0 = v0 = 28 m/s
v=0
Car obviously stops when crash ends! 
Wanted unknown: t.
But we don’t know acceleration a or
distance x either! Estimate x = 1.0 m
This has to be a 2 step problem! First, use
(2) to solve for a: 0 = (v0)2 + 2a(x – 0) so
a = - (v0)2∕(2x) = - (28)2 ∕(2) = - 390 m/s2
(1)
This is a HUGE acceleration!!
Now, use (1) to solve for t: 0 = v0 + at so
(2)
(3)
t = - (v0) ∕a = 0.07 s !!!
(4)
Example: Estimate Breaking Distances

v = v0 = constant = 14 m/s
t = 0.50 s, a = 0, x = v0t = 7 m

a = - 6.0 m/s2, x0 = 7 m
v decreases from 14 m/s to zero
Example: Estimate Breaking Distances

v = v0 = constant = 14 m/s
t = 0.50 s, a = 0, x = v0t = 7 m
Note: The 2nd time interval is the
actual braking period when the
car slows down & comes to a stop.
Stopping distance depends on
1) the driver’s reaction time,
2) the car’s initial speed,
3) the car’s acceleration.

a = - 6.0 m/s2, x0 = 7 m
v decreases from 14 m/s to zero
Example: Estimate Breaking Distances

v = v0 = constant = 14 m/s
t = 0.50 s, a = 0, x = v0t = 7 m
Note: The 2nd time interval is the
actual braking period when the
car slows down & comes to a stop.
Stopping distance depends on
1) the driver’s reaction time,
2) the car’s initial speed,
3) the car’s acceleration.

a = - 6.0 m/s2, x0 = 7 m
v decreases from 14 m/s to zero
v0 = 14 m/s, v = 0
v2 = (v0)2 + 2a(x – x0)
x = x0 + [v2 - (v0)2]/(2a)
x = 7 m + 16 m = 23 m
Example: Braking distances continued
v = const.

v = v0 + at

Plots for this case:
v(t)
Velocity vs time v(t)
x = x0 + v0t + (½)at2
x(t)
Position vs time x(t)
x = v0t
Example: Fastball
Known: x0 = 0, x = 3.5 m, v0 = 0, v = 44 m/s
Wanted: a
Example: Fastball
Known: x0 = 0, x = 3.5 m, v0 = 0, v = 44 m/s
Wanted: a
Use: v2 = (v0)2 + 2a (x - x0)
 a = (½)[v2 - (v0)2]/(x - x0) = 280 m/s2 !
Example: 2 Moving Objects: Police & Speeder
A car, speeding at v0S = 150 km/h (42 m/s) passes a still police car (v0P =
0) which immediately takes off (accelerates!) in hot pursuit. Using simple
assumptions, such as that the speeder continues at constant speed v0S = 42
m/s (& also for the acceleration aP of the police car!), ESTIMATE how
long it takes the police car to overtake the speeder. Then ESTIMATE the
police car’s speed at that moment & decide if the assumptions were
reasonable.
Note! Before working this problem, we need to work another problem,
which will give us an ESTIMATE of the acceleration aP of the police car.
In order to do this, we take numbers from ads for the type of car the police
drive. These claim that this car can accelerate from rest to 100 km/h (28
m/s) in 5.0 s. Using v = v0 + aPt with these numbers gives 28 = 0 + aP(5) or
aP = 5.6 m/s2. So, to solve this problem of the police car catching up to the
speeder, we use this ESTIMATE for the acceleration aP
Problem now restated is: A car, speeding at v0S = 150 km/h (42
m/s) passes a still police car (v0P = 0) which immediately takes off
(accelerates!) in hot pursuit. Assume that the speeder continues at
constant speed v0S = 42 m/s & that aP = 5.6 m/s2. ESTIMATE how
long it takes the police car to overtake the speeder. Then ESTIMATE the
police car’s speed at that time & decide if the assumptions were reasonable.
Solution: The speeder moves at constant speed v0S = 42 m/s so, at
some time t later it has moved a distance xS = v0St. In that same time
t the police car has moved a distance xP = (½)aPt2 When the police
car catches the speeder, the two distances must be the same. So, we
equate them and solve for t: xS = v0St = xP = (½)aPt2. This is a
quadratic equation for t, which has solutions; t = 0 & t = 15 s.
The Problem also asks: ESTIMATE the police car’s speed at that
time (t = 15 s) & decide if the assumptions were reasonable.
Use :
vP = v0P + aPt
Gives:
vP = 84 m/s (300 km/h ≈ 190 mph!)
Not only unreasonable, but also very dangerous!
For the assumptions we’ve made, the x versus t & v versus
t curves are shown here:
More reasonable v
versus t curves are:
Example: Carrier Landing
A jet lands on an aircraft carrier at velocity
v0 = 140 km/h (63 m/s).
a) Calculate the acceleration (assumed constant) if it stops
in t = 2.0 s due to the arresting cable that snags the
airplane & stops it.
b) If it touches down at position x0 = 0, calculate
it’s final position.
Example: Carrier Landing
A jet lands on an aircraft carrier at velocity
v0 = 140 km/h (63 m/s).
a) Calculate the acceleration (assumed constant) if it stops
in t = 2.0 s due to the arresting cable that snags the
airplane & stops it.
b) If it touches down at position x0 = 0, calculate
it’s final position.
Solutions
a) v0 = 63 m/s, t = 2.0 s = time to stop.
When it is stopped, v = 0. So, use v = v0 + at = 0, which gives
a = - (v0/t) = - (63/2) = -31.5 m/s2
Example: Carrier Landing
A jet lands on an aircraft carrier at velocity
v0 = 140 km/h (63 m/s).
a) Calculate the acceleration (assumed constant) if it stops
in t = 2.0 s due to the arresting cable that snags the
airplane & stops it.
b) If it touches down at position x0 = 0, calculate
it’s final position.
Solutions
a) v0 = 63 m/s, t = 2.0 s = time to stop.
When it is stopped, v = 0. So, use v = v0 + at = 0, which gives
a = - (v0/t) = - (63/2) = -31.5 m/s2
b) Use x = x0 + v0t + (½)at2 , which gives
x = x0 + v0t + (½)at2 = 0 + (63)(2) + (½)(-31.5)(2)2
x = 63 m