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 slightlyt1 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 tv2 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 speedof 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 ofthe 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 !