Sect. 2-5: Motion at Constant Acceleration
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Transcript of Sect. 2-5: Motion at Constant Acceleration
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 !