PHY183-Lecture6

# PHY183-Lecture6 - Physics for Scientists & Engineers 1...

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Unformatted text preview: Physics for Scientists & Engineers 1 Spring Semester 2006 Lecture 6 January 20, 2006 Physics for Scientists&Engineers 1 1 Acceleration Vector Average acceleration = change of velocity during a given time interval Dv aDt = Dt Instantaneous acceleration vector Dv dv a (t ) = lim aDt = lim Dt 0 Dt 0 Dt dt Acceleration vector is the second derivative of the position vector d d d ^ d2 a (t ) = v = x ~ = 2 x dt dt dt dt January 20, 2006 Physics for Scientists&Engineers 1 2 Example: Velocity & Acceleration 2 Graph of x(t) = 17.2 m - (10.1 m)(t / s)+(1.1 m)(t /s) and v(t) = -10.1 m/s+(2.2 m/s)(t / s) Acceleration: January 20, 2006 a(t) = 2.2 m/s2 Physics for Scientists&Engineers 1 3 Example: 100 m Sprint Carl Lewis' World Record, 1991 World Championship vi f Dx x f - xi = = Dt t f - ti ai f Dv v f - vi = = Dt t f - t i Fit: v = 11.58 m/s January 20, 2006 Physics for Scientists&Engineers 1 4 Quiz Question #1 e b c a d At which segment(s) is the acceleration negative? A) B) C) D) a-c c-d c-e d-e January 20, 2006 Physics for Scientists&Engineers 1 5 Quiz question #2 e b c a d At which point(s) does the acceleration equal zero? A) B) C) D) E) a b c d e 6 January 20, 2006 Physics for Scientists&Engineers 1 Inverse Relationships Given velocity as function of time v(t)=dx/dt, find position by integration: dx dx v= dt fi vdt = t 0 dt dt = dx = x - x0 x(t ) = x0 + v(t ')dt ' Position at t=0 is x0 Similar integration to obtain the velocity from the acceleration a(t)=dv/dt v(t ) = v0 + a (t ')dt ' 0 January 20, 2006 Physics for Scientists&Engineers 1 7 t Linear motion with Constant Acceleration What is that about? A lot of cases we have already looked at ! Use the integral formula and set a = constant: v(t ) = v0 + a (t ')dt ' = v0 + a dt ' = v0 + at fi v(t ) = v0 + at 0 0 t t The velocity is linear in time Integrate velocity to find position: x(t ) = x0 + v(t ')dt ' = x0 + (v0 + at ')dt ' 0 0 t t = x0 + v0 January 20, 2006 t 0 x(t) = x0 + v0t + 1 at 2 dt '+ a t ' dt ' fi 2 0 Physics for Scientists&Engineers 1 8 t The position is quadratic in time Position and Average Velocity Start from definition v 0t Dx x(t) - x 0 x(t) - x 0 = = = Dt t -0 t Now multiply both sides with t and add x0. We get x(t ) = x0 + v0t t (valid in general, not just for constant a.) January 20, 2006 Physics for Scientists&Engineers 1 9 Average Velocity If velocity depends linearly on time, what is the average velocity over the time interval from t0 to t? Answer: v0t v v(t) 1 t 1 t = v(t ')dt ' = (v0 + at ')dt ' v0 t 0 t 0 v0 t a t = dt '+ t ' dt ' = v0 + 1 at t 2 0 t t 0 t 0 = 1 v0 + 1 (v0 + at ) 2 2 So, between 0 and t: = 1 (v0 + v(t )) 2 V(t) v0t = 1 [v0 + v(t )] 2 (valid only for constant a.) Physics for Scientists&Engineers 1 10 January 20, 2006 Relation between x and v Solve v (t ) = v0 + at for the time and get: v(t ) - v0 t= a Substitute this result into the position expression: x(t ) = x0 + v0t + 1 at 2 2 v(t ) - v0 ^ 1 v(t ) - v0 ^ = x0 + v0 ~ + 2 a ~ a a 2 2 v(t )v0 - v0 1 v(t ) 2 + v0 - 2v(t )v0 = x0 + +2 a a Subtract x0 from both sides and multiply with a: January 20, 2006 Physics for Scientists&Engineers 1 11 2 Expression for v2 (cont.) 2 2 v(t )v0 - v0 1 v(t ) 2 + v0 - 2v(t )v0 x(t ) = x0 + +2 a a 2 2 a[ x(t ) - x0 ] = v(t )v0 - v0 + 1 [v(t ) 2 + v0 - 2v(t )v0 ] 2 2 a[ x(t ) - x0 ] = 1 v(t ) 2 - 1 v0 2 2 Result: 2 v(t ) 2 - v0 = 2a[ x(t ) - x0 ] January 20, 2006 Physics for Scientists&Engineers 1 12 Summary: Five Kinematical Equations One-dimensional motion with constant acceleration: v(t ) = v0 + at x(t ) = x0 + v0t + 1 at 2 2 v0t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0t t v(t ) - v = 2a[ x(t ) - x0 ] Can solve practically all one-dimensional problems January 20, 2006 Physics for Scientists&Engineers 1 13 2 2 0 Example: Airplane Take-off (1) Experiment: using strain gauge, measure acceleration during airplane take-off Result: Constant acceleration, to good approximation a = 4.3 m/s2 January 20, 2006 Physics for Scientists&Engineers 1 14 Example: Airplane Take-off (2) Question 1: Assuming constant acceleration of a = 4.3 m/s2, starting from rest, what is the takeoff speed of the airplane at t1 =18 seconds? Answer 1: Starting from rest: initial velocity v0 = 0. t1 = 18s a = 4.3ms -2 v ( t1 )? v0 = 0 v(t ) = v0 + at x(t ) = x0 + v0t + 1 at 2 2 v0t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0t t 2 v(t ) 2 - v0 = 2a[ x(t ) - x0 ] v(t1 ) = v0 + at1 = 0 + (4.3 m/s2 )(18 s) = 77.4 m/s (175 mph) January 20, 2006 Physics for Scientists&Engineers 1 15 Example: Airplane Take-off (3) Question 2: How far down the runway has this airplane moved by the time it takes off? v(t ) = v0 + at Answer 2: x0 = 0 v0 = 0 a = 4.3m.s t1 = 18s -2 x(t ) = x0 + v0t + 1 at 2 2 x ( t1 )? v0t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0t t 2 v(t ) 2 - v0 = 2a[ x(t ) - x0 ] x(t1 ) = x0 + v0t1 + 1 at12 = 1 (4.3 m/s2 )(18 s)2 2 2 = 697 m (2,290 ft) January 20, 2006 Physics for Scientists&Engineers 1 Main runway at Lansing airport is 7500 ft long 16 Race of Two Balls Two balls roll down incline to gain speed; after that: Ball 1 rolls on horizontal track Ball 2 rolls on "roller coaster" track Question 1 (multiple choice): A. Ball 1 will arrive first B. Ball 2 will arrive first C. Both balls will arrive at the same time Question 2 (multiple choice): A. Ball 1 will arrive with higher speed B. Ball 2 will arrive with higher speed C. Both balls will arrive with the same speed January 20, 2006 Physics for Scientists&Engineers 1 17 Race of Two Balls, quantitative Let's look at accelerations and velocities a(x) v(x) x a(x) v(x) x January 20, 2006 Physics for Scientists&Engineers 1 18 ...
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## This note was uploaded on 03/19/2008 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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