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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 .
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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 motiononly!): 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 VALIDUNLESS a = CONSTANT!!! Usually x0 = 0. Sometimes v0 = 0
All we need for 1 dimensional constant-acceleration problems: NOT VALIDUNLESS a = CONSTANT!!!
Physics and Equations IMPORTANT!!! • Even though these equations & their applications are important, Physics is nota 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 • Read the whole problem.Make sure you understand it. Read it again. • Decide on the objects under study & what the time interval is. • Draw a diagram & choose coordinate axes. • Write down the known(given) quantities, & the unknown ones needed. • What physics applies?Plan an approach to a solution. • Which equationsrelate known & unknown quantities? Are they valid in this situation? Solvealgebraicallyfor the unknown quantities, & check that your result is sensible (correct dimensions). • Calculate the solution, round it to appropriate number of significant figures. • Lookat the result - is it reasonable?Does it agree with a rough estimate? • Check the units again.
Bottom Line: THINK! DO NOTBLINDLY APPLY FORMULAS!!!!
Example 2-6: Runway Design Solutions (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 thismeans 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! 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 (1) (2) (3) (4)
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? 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! Obviously, it starts from rest!!
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 accelerationa or distance x either! Estimatex = 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 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/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 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 v = const. Plots for this case: Velocity vs time v(t) Position vs time x(t) v = v0 + at v(t) x = x0 + v0t + (½)at2 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 !
