working.kinematics.problems

working.kinematics.problems - n d 6 WaYA`FWVTgABR hbAUcB...

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9`hRQ9aP2ciA89hRlrTB9BA6ebcia8fQEpaDQ9EDfc cQaPb APQXPW` `AGBdADEW9Bg6VaPA6q9BADEW9Bg6EYcfhRlEDfcd8 EABcAPQq` 9Be6 EDD hB9`hR9`dR8 AP QlV8D iP h8F`e6P eWqEPclfEDAB` cPb AP QF2P cfQQdP qDVQ9R R Wc rfP B W W Y 8 P i D B 8 6APWgABP QaW iA88kaPc8b 9iEYccRY [email protected] [email protected] EP GB` eRQTWP caYb 9`GRaDVVASG8jFBA6R [email protected] QjcBB 9oR AD9DhRQViAUTWaY9`RD kx8S cBB 9W 7QB 9d ADQ SW B U ` aP` B 8 TiEDP h8hB9`9`FDaR9`EDk9BR QQfg QqAUB lVP [email protected] A82eRD QdW 9T6R GB)aPQcQilmcG8YPPD`6 g8 8 caiD m)aPE9B` 9ia6EY9`cEDB [email protected] Q8B 92A6A6C88GRQEDD iaP B V9`Uh8 S FDED9B9`QcgBSD6 ABQiciXR9`PWr h8 A8ilQQ9BWeRDG8U 9R 9iEYljGRY fri A8 @ d [email protected]FDEDgBEYcARcxaDhRA`AhaUAU9WEPcErB A6aT`[email protected] D GiADW U 6S B R ` D8 R @ UP DP 8PiP B R 8 W @ U D R BP 8P ` R d Bc9Qic9hR2EABhiG82pG8EcihbF`EDcl9Be69DADaWTU7aDQ9XPD d hBVAgF8QjQiyTrV2ccQ T6Er aPGB7cQxiSPDR` a89`eRg8QhEؤBW [email protected])hS 8 EUcc9BB R R R8 ` U D Bg R BP B W Y ` R ` U 6S 68 DP 8PiP WY s EXAMPLE 2-9 You see a stalled vehicle in front of you, apply the brakes, and decelerate at a constant rate of 5 m/s2. What's your stopping distance if your initial speed is (a) 15 m/s (about 34 mph), or (b) 30 m/s? your car, initially stalled car -x +x STEP 1: define the coordinate system. I choose the origin at your initial position and the x-axis is along the initial velocity. STEP 2: x(t=0) = 0 (your initial position) v(t=0) = v0 = 15 m/s (or 30 m/s, or whatever it may be...) a(t) = -a0 = -5 m/s2 STEP 3: v(t) = a(t)dt = -a0t + C1 v(t=0) = C1 = v0 by the initial condition hence v(t) = -a0t + v0 x(t) = v(t)dt = - a0t2 + v0t + C2 x(t=0) = C2 = 0 by the initial condition hence x(t) = -a0t2 + v0t STEP 4: The condition of stopping at a time, call it t=tstop, implies v(tstop) = -a0tstop + v0 = 0 The stopping distance is the position at the time tstop x(tstop) = - a0tstop + vtstop = D (stopping distance) This gives us two equations in two unknowns (D and tstop) STEP 5: solve by algebra! second equation tstop = v0/a0 first equation D = -a0(v0/a0)2 + v0(v0/a0) v02 D= 2a0 STEP 6: check the units: left side is in meters right side is (m2/s2)/(m/s2) is also meters! if v0 (your initial speed) is bigger, then stopping distance D is bigger...this is as you expect! if your deceleration a0 is bigger, then stopping distance D is shorter...this is also as you expect! STEP 7: put in a0 = 5 m/s2, v0 = 15 m/s D = 45 m put in v0 = 30 m/s D= 180 m Notice how easy it is to solve for different sets of numbers, if you've worked the problem in letters! EXAMPLE 2-12 A car going 100 km/hr hits a wall. What is the acceleration experienced by a passenger sitting 0.75 m from the front bumper? Assume constant deceleration. place of collision at origin -x passenger initially at x0 = -0.75 m +x STEP 1: Choose coordinates so that the origin is the place of the collision, and the initial position of passenger is at x = x0 = 0.75 m. STEP 2: x(t=0) = 0 v(t=0) = v0 = 27.8 m/s (100 km/hr) a(t) = a0 (acceleration is not known---assign it a letter and move on! STEP 3: v(t) = a0 dt = a0t + C1 v(t=0) = C1 = v0 therefore v(t) = a0t + v0 x(t) = (a0t + v0) dt = a0t2 + v0t + C2 x(t=0) = C2 = x0 = 0.75 m therefore x(t) = a0t2 + v0t x0 STEP 4: We know that the passenger ends up at the place of collision hence x(tend) = a0tend2 + v0tend x0 = 0 We also know the passenger is stationary at the end hence v(tend) = a0tend + v0 = 0 We have two equations in two unknowns (a0 and tend)! STEP 5: Solving by algebra: second equation gives tend = v0/a0; plugging into the first equation and solving for a0 gives: a0 = v02/2x0 STEP 6: The units check out as in example 2-9 (meters on both sides) increasing the car's initial speed, v0, means that the deceleration experienced will be greater, as we expect. Decreasing the distance from the bumper to the passenger also increases the resulting deceleration, as we expect. STEP 7: Plugging in numbers: a0 = (27.78)2/(2)(0.75) = 514 m/s2 same answer as the book gets EXAMPLE 2-13 A student throws a cap in the air with initial speed 14.7 m/s. It rises and falls. Find the time it takes to reach its highest point and the height H of the top of its trajectory. Assume constant acceleration of g. STEP 1: hat initially at origin STEP 2: y(t=0) = 0 v(t=0) = v0 = 14.7 m/s a(t) = g = 9.81 m/s2 STEP 3: v(t) = g dt = gt + C1 hat initially at origin +y v(t=0) = C1 = v0 hence v(t) = gt + v0 y(t) = ( gt + v0) dt = gt2 + v0t + C2 y(t=0) = C2 = 0 hence y(t) = gt2 + v0t y STEP 4: at the top of its trajectory, v(ttop) = gttop + v0 = 0 the y-value at this point is the desired height: y(ttop) = gttop2 + v0ttop = H We have two equations in two unknons (ttop and H)! STEP 5: solve: ttop = v0/g, H = v02/2g STEP 6: units check out. Increasing v0 makes H bigger as expected STEP 7: plugging in: H = (14.7)2/(2)(9.81) = 11 m EXAMPLE 2-14 Car A travels 25 m/s (90 km/hr). Car B starts from rest just as it is passed by car A and accelerates at a constant rate of 5.0 m/s2. When does car B catch car A? How fast is car B traveling at that point? car A passes car B at origin -x +x STEP 1: Choose coordinates so that the origin is the place where car A passes car B initially. Lay the x-axis along the velocities. STEP 2: This time we have two objects car A: xA(t=0) = 0 vA(t=0) = v0 =25 m/s aA(t) = 0 car B: xB(t=0) = 0 vB(t=0) = 0 aB(t) = a0 = 5 m/s2 STEP 3: vA(t) = xA(t) = 0 dt = C1 = v0 v0 dt = v0t + C2 xA(t=0) = C2 = 0 hence xA(t) = v0t vB(t) = a0 dt = a0t + C3 vB(t=0) = C3 =0 hence vB(t) = a0t xB(t) = a0t dt = a0t2 + C4 xB(t=0) = C4 =0 hence xB(t) = a0t2 STEP 4: The condition for car B to catch car A is that their positions at that time are equal: xA(tpass) = xB(tpass): a0tpass2 = v0tpass This is one equation in one unknown (tpass)! The speed of car B at this time is vB(tpass) = a0tpass STEP 5: tpass = 2v0/a0 vB(tpass) = 2v0 STEP 6: units check out! If v0 is bigger, it takes car B longer to pass, as expected. If car B's acceleration (a0) is bigger, it takes car B less time to pass, as expected. STEP 7: tpass = (2)(25)/5 =10 s, vB(tpass) = (2)(25) = 50 m/s (same answers as the book gets) EXAMPLE 2-15 How fast is car B traveling when it's 25 m behind car A? STEPS 1-2-3: same as above! STEP 4: The condition that car B is D=25 m behind is xB(t) = xA(t) D a0t2 = v0t D This is one equation in one unknown (t)! The speed of car B at this time is vB(t) = a0t STEP 5: we need to solve the quadratic equation for t. There are two solutions: t = [v0 + v02 2a0D ]/a0 If you study closely (or put in numbers) you notice that the minus solution is a negative number. We don't want this solution, we want the other one which corresponds to the time after t=0 when car B is D=25 m behind car A. Therefore use the + solution to calculate vB(t) = v0 + v02 2a0D STEP 6: check the units: vB(t) is m/s, term v0 obviously is m/s second term under radical is (m/s2)(m), which when square-rooted is also m/s. Units are correct! If you make D smaller, vB(t) is bigger, as expected. (Making D=0 recovers the same solution as the last example, as expected!) STEP 7: vB(t) = (25) + (25)2 (2)(5)(25) = 44.36 m/s EXAMPLE 2-16 An elevator accelerates upward at a rate of 4.0 m/s2. Its ceiling is 3.0 m high. A screw falls out of the ceiling (accelerating downward at g. How long does it take the screw to hit the bottom of the elevator? STEP 1: elevator floor initially at origin +y elevator initially at origin screw initially at h=3 m STEP 2: There are two objects. For the screw (S): yS(t=0) = y0 = 3.0 m vS(t=0) = v0E = not given! aS(t) = g = 9.81 m/s2 For the elevator (E) yE(t=0) = 0 vE(t=0) = v0E = not given! y aE(t) = a0 = +4.0 m/s2 The screw and elevator have the same initial speed, because the screw is on the elevator to start with. We are not given this speed, but we assign it a letter (v0E) and move on! STEP 3: vS(t) = g dt = gt + C1 vS(t=0) = C1 = v0E hence vS(t) = gt + v0E yS(t) = (gt + v0E) dt = gt2 + v0Et + C2 yS(t=0) = C2 = y0 hence yS(t) = gt2 + v0Et + y0 STEP 3 (Continued): vE(t) = a0dt = a0t + C3 vE(t=0) = C3 = v0E hence vE(t) = a0t + v0E yE(t) = (a0t + v0E) dt = a0t2 + v0Et + C4 yE(t=0) = C4 = 0 hence yE(t) = a0t2 + v0Et STEP 4: The condition that the screw hits the elevator floor is that their positions become equal: yS(thit) = yE(thit) plugging in from step 3: gthit2 + v0Ethit + y0 = a0thit2 + v0Ethit The unknown v0E cancels out! You are left with one equation in one unknown (thit). STEP 5: solve: thit2 = 2y0/(g + a0) STEP 6: units check out: left side is (seconds)2, right side is (m)/(m/s2) which is also (seconds)2. Increasing the initial height of the screw, y0, lengthens the time for it to fall, as expected. Increasing the acceleration of the elevator, a0, shortens the time, as expected. STEP 7: plugging: thit= (2)(3)/(9.81 + 4.0) = 0.66 s EXAMPLE 2-17 Suppose the elevator is initially moving upward with v0E=16 m/s the instant when the screw falls. How far does the elevator rise while the screw is falling? What is the displacement of the screw during its fall? What are the velocity of the screw and the elevator when they impact? STEPS 1-2-3-4-5 are the same as for the previous example! Once you've calculated thit, you can now trivially calculate yE(thit) = a0thit2 + v0Ethit (distance elevator rises during screw's fall) displacement of screw during fall = y0 yS(thit) = y0 [gthit2 + v0Ethit + y0] = gthit2 v0Ethit vE(thit) = a0thit + v0E vS(thit) = gthit + v0E (velocity of elevator when screw hits it) (velocity of screw when it hits elevator floor) When you plug in numbers for these, you get the same answers as on p 47 of the book. EXAMPLE 2-11 An electron starts from rest, accelerates with 5.33x1012 m/s2 for 0.150 sec, drifts with constant speed for 0.200 sec, and slows to a stop with an acceleration of -2.67x1013 m/s2. How far does it travel? electron initially at origion -x +x STEP 1: define the coordinate system. I choose the origin at the electron's initial position and the x-axis is along the acceleration. STEP 2: there is only one object, the electron. Its initial conditions are: x(t=0) = 0 v(t=0) = 0 But the acceleration is a little trickier to write: a1 for t < t1 = 0.15x10-6 sec a(t) = 0 for t1 < t < t2 = 0.350x10-6 sec -a2 for t > t2 12 m/s2 and a = 2.67x1013 m/s2 where a1 = 5.33x10 2 STEP 3: integrate a(t) to get v(t). You have to break the integral up into three pieces: t1 for t < t1, v(t) = a1 dt = a1t 0 t1 t for t1 < t < t2, v(t) = a1 dt + 0 dt = a1t1 0 t1 for t > t2, t t2 t1 v(t) = a1 dt + 0 dt + -a2 dt = a1t1 a2 (t t2) 0 t2 t1 integrate v(t) to get x(t); similar story, but we are only interested in the position at the end (when t > t2), so I only calculate that case: t1 t2 t x(t) = a1t dt + a1t1 dt + [a1t1 a2 (t t2)] dt 0 t1 t2 x(t) = a1t12 + a1t1(t2 t1) + (a1t1 + a2t2)(t t2) a2(t2 t22) STEP 4: the electron is at rest at the end, which means v(tend) = a1t1 a2 (tend t2) = 0 Furthermore the electron is at position D at the end which means x(tend) = a1t12 + a1t1(t2 t1) + (a1t1 + a2t2)(tend t2) a2(tend2 t22) = D This is two equations in two unknowns (D and tend)! STEP 5: Solve the first equation for tend = (a1t1 + a2t2)/a2 = (a1/a2)t1 + t2 This can be substituted into the second equation to calculate D. STEP 6: Note that when the electron's initial acceleration, a1, gets larger, tend gets larger, and when the electron's deceleration, a2, gets larger, tend gets smaller---both in agreement with intuition. ...
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