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In this assignment we will go through the steps to conduct a basic demand analysis. For this purpose, we will use scanner data obtained from Kraft (now Kraft Heinz). We only use a small sub…Full description
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EGR 232 Dynamics: Homework Set 02 Fall 2012 Problem 11.34 A truck travels 220 m in 10 s while being decelerated at a constant rate of 0.6 m/s2. Determine a) its initial velocity, b) its final velocity c) the distance traveled during the first 1.5 s. ___________________________________________________________________________ Solution: a = -0.6 m/s2 x = 220 m x0 = 0m t = 10 s t0 = 0 s This is a constant acceleration problem. Principle: v v0 at
x x0 v0t
1 2 at 2
x x0
1 (v v0 )t 2
v 2 v02 2a(x x0 )
Initial velocity may be found using:
1 x x0 v0t at 2 2
v0
x x0 t
220m 0m
1 2 at 2
1 (0.6m / s2 )(10s)2 2 25m / s 10s
Final velocity: v v0 at 25m / s (0.6m / s2 )(10s) 19m / s Distance after 1.5 seconds:
x x0 v0t
1 2 1 at 0m (25m / s)(1.5s) (0.6m / s2 )(1.5s)2 = 36.83 m 2 2
EGR 232 Dynamics: Homework Set 02 Fall 2012 Problem 11.37: A sprinter in a 100 m race accelerates uniformly for the first 35 m and then runs with a constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine a) his acceleration, b) his final velocity, c) his time for the race. ___________________________________________________________________________ Solution: from A to B: Uniform acceleration from B to C: Uniform velocity: xC = 100
xB = 35 m tA-B = 5.4 s
xA = 0
vC=vB vA = 0 For uniform acceleration: v v0 at
x x0 v0t
1 2 at 2
1 v 2 v02 2a(x x0 ) (v v0 )t 2 x x0 vt For uniform velocity: x x0
Acceleration from A to B: x x0 v0t 35m 0 (0)(5.4) 1 2.4m / s2 xB x A v Ata b ata b2 a 1 2 1 2 2 t (5.4s) 2 2 therefore his velocity at 35 m will be: v B v A at a b 0 (2.4m / s2 )(5.4s) 12.96m / s therefore final velocity is 12.96 m/s
Time to get from B to C. xC xB vBtb c
t
xC xB (100 35)m 5.015s vB 12.96m / s
Total Time: ttotal t a b tb c 5.4 5.015 10.42s
EGR 232 Dynamics: Homework Set 02 Fall 2012 Problem 11.45 Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a constant speed of 60 mi/h. At t = 0, A starts and accelerates at a constant rate aA, while at t = 5 s, B begins to slow down with a constant deceleration of magnitude aA/6. Knowing that when the cars pass each other x = 294 ft and vA = vB, determine a) the acceleration aA, b) when the vehicles pass each other, c) the distance d between the vehicles at t = 0. ------------------------------------------------------------------------------------------------------------------Solution: At position 1: t = 0: xA = 0 vA = 0 aA = constant vB = -60 mi/hr = -88 ft/s constant xB = d At position 2: t = 5: aB = -aA/6 = constant At position 3: xA = xB = 294 ft velocities are equal: vA3 = -vB3
time from 1 to 2: 5 s time from 2 to 3: t – 5
Principles: Uniform acceleration v v0 at
x x0
and uniform velocity
1 x x0 v0t at 2 2
1 (v v0 )t 2
x x0 vt
v 2 v02 2a(x x0 )
Equations of Motion for Car A between position 1 and 3: 1 v v0 at x x0 v0t at 2 2 1 v A3 aAt 294 (aA )t 2 2 Equation of Motion of Car B between positions 1 and 2: xB2 xB1 vBt xB2 d (88ft / s)(5s) xB2 d 440ft
Equation of Motion of Car B between positions 2 and 3: v v0 at vB3 88 (aA / 6)(t 5)
x x0 v0t
1 2 at 2
294 xB2 (88)(t 5)
1 (aA / 6)(t 5)2 2
Solve the set of 6 equations identified on the previous page: Note that there are 6 unknowns: vB3, vA3, xB2, aA, t, and d v A3 v B3 v A3 aAt
294
1 (aA )t 2 2
xB2 d 440
vB3 88 (aA / 6)(t 5)
294 xB2 (88)(t 5)
1 (aA / 6)(t 5)2 2
Start by eliminating xB2 and vB3 v A3 aAt v A3 88 (aA / 6)(t 5)
294
1 (aA )t 2 2
1 (aA / 6)(t 5)2 2 next get rid of vA3 and notice that d is only present in the final equation, so solve it for d. a t 5a 7 5 aAt 88 (aA / 6)(t 5) aAt 88 A A aAt 88 aA 6 6 6 6 75.42 5 1 t 294 (aA )t 2 aA 7 2 294 (d 440) (88)(t 5)
1 (aA / 6)(t 5)2 2 combine the first equations to solve for aA: 1 75.42 1 107.67 5689.4 294 (aA )( 0.714)2 (aA )(0.5102 ) 2 aA 2 aA aA2 d 294 440 (88)(t 5)
588aA 0.5102aA2 107.67aA 5685.2
0 0.5102aA2 480.33aA 5685.2
(solve with quadratic equation)
Since 929.46 is very large, the correct acceleration value is aA =11.98 ft/s2 which give a total travel time of 75.4 t 0.714 7.008s 11.98 the distance d is
Another way to solve this set of equations without the lengthy manual solution is to use the equation solving ability of a program such as Mathematica.
EGR 232 Dynamics: Homework Set 02 Fall 2012 Problem 11.49 The elevator shown in the figure moves downward with a constant velocity of 15 ft/s. Determine a) the velocity of the cable C, b) the velocity of the counterweight W, c) the relative velocity of the cable C with respect to the elevator, d) the relative velocity of the counterweight W with respect to the elevator. --------------------------------------------------------------------------------------------------------------------Solution: (assume down is + direction) vE = 15 ft/s Constrained motion and uniform velocity: For Rope L1: L1 xC xE xE For Rope L2: L2 xE xW
xC
xE
L1
L2
L1 xC 2 xE
L2 xW xE
Taking the derivative of each of the constrained motion equations: dL1 dxC dx dL2 dxW dxE 2 E dt dt dt dt dt dt 0 vW vE 0 vC 2vE vC 2vE
vW v E
Since vE 15ft / s (downward)
then vC 2vE 2(15) 30ft / s 30ft / s upward
and vW vE (15) 15ft / s 15ft / s upward
Relative velocity of C with respect to the elevator: vC / E vC vE 30ft / s 15ft / s 45ft / s 45ft / s downward Relative velocity of W with respect to the elevator: vW / E vW v E 15ft / s 15ft / s 30ft / s 30ft / s downward