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 Strategies1. 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 physicsphysics 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 DesignYou’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 & UnknownsSolutions(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/sNote that this means take the square root! That 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!

(1)

(2) (3)

(4)

Example 2-7: Acceleration of a Car

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!

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!!

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?

Example 2-8: Air Bags

Known: x0 = v0 = 28 m/s v = 0Car obviously stops when crash ends! Wanted unknown: t. But we don’t know acceleration a or

distance x either! Estimate x = 1.0 mThis 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

This is a HUGE acceleration!!

Now, use (1) to solve for t: 0 = v0 + at so

t = - (v0) ∕a = 0.07 s !!!

(1)(2)(3)

(4)

Example 2-9: Estimate Breaking Distances

v = v0 = constant = 14 m/st = 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

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

Note: The 2nd 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.

Example 2-9: Braking distances continued

x(t)

v(t)

v = const.

v = v0 + at

x = v0t

x = x0 + v0t + (½)at2

Plots for this case:

Velocity vs time v(t)

Position vs time x(t)

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 !