Sect. 2-5: Motion at Constant Acceleration
Motion with Constant Acceleration
• 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 case ONLY.
– The kinematic equations for uniform
acceleration in one dimension.
Constant Acceleration
•
Derivation is in the text. Also done in class. Also on next slide! Read on your own!
• In the derivation, its useful to change notation slightly
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)
• Using these, 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)
Derivation on the board!
Constant Acceleration Equations
• Results (one dimensional motion only!):
v = v0 + at
(1)
x = x0 + v0 t + (½)a t2
(2)
v2 = (v0)2 + 2a (x - x0)
(3)
v = (½) (v + v0)
(4)
NOT VALID UNLESS a = CONSTANT!!!
Usually 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. Read it again.
2. Decide on the objects under study & what the time interval is.
3. Draw a diagram & choose coordinate axes.
4. Write down the known (given) 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 2-6: 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?
Solutions
Table of Knowns & Unknowns
(a) Use Eq. (3):
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 matters!
(b) Use Eq. (3) again with
v = vmin = 27.8 m/s. Solve for
(1)
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!
(2)
(3)
(4)
Example 2-7: 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 matters!
Example 2-8: 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 2-9: Estimate Breaking Distances

v = v0 = constant = 14 m/s
t = 0.50 s, a = 0, x = v0t = 7 m
2nd
Note: The
time interval is the
actual braking period when the car
slows down & comes to a stop.
Stopping distance depends on 1)
driver’s reaction time, 2) car’s
initial speed, 3) car’s acceleration.
2
a = - 6.0 m/s , 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 2-9: 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 (5th Edition), 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 !