Kinematics (1D)

The study of the motion of objects, and the related concepts of force and energy, form the field entitled Mechanics. This field, Mechanics, is divided into two parts: Kinematics, which relates to how objects move, and Dynamics, which describes why objects move(usually dealing with force).

This section will deal with translational motion.

The derivative of a position function yields the function of velocity. The second derivative yields the function of acceleration. That is if s(t) is the position function at time t, then v(t) = s'(t) and a(t) = s''(t)

Problem 1: Displacement.
A duck decides to walk his way home from school. In order to do so, the duck must walk 200 meters west from school. What is the displacement?
/\x=x2-x1. Change in position x. Let x1 be initial starting point, x2 the final.
/\x=200m-0.
Total displacement = 200m.

Problem 2: Average Speed
It takes a duck 10 seconds to fly 50 meters. What is its average speed?
Average speed = distance//\time.
Average speed = 50m/10s.
Average speed = 5m/s.

Problem 3: Average Velocity
Unlike speed, velocity is a vector, meaning velocity has magnitude and direction. It takes a duck 10 seconds to fly west 50m then east 30m. What is its average velocity?
Average velocity = displacement//\time. consider west as -x, east as +x.
Average velocity = 30m-50m/10s
Average velocity = -2m/s

Problem 4: An angry duck takes upon the position function of s(t) = 5t^2 10t+3 m. Find its velocity and acceleration at t = 10s.

v(t) = s'(t) = 10t+10. At t = 10, v = 10(10) +10 = 110m/s
a(t) = s''(t) = v'(t) = 10m/s^2

Problem 5: A dumbchicken at (0,0) is planning on getting to a barbeque(HAHAHAHAHAHA) at (210,800) but there seems to be some wind blowing at a constant 50m/s from the east. What is the chicken's actual speed and where does he end up after 10 seconds if he's flying at 100m/s at a direction of i+j?

Here we want to find the actual velocity vector, V_actualchicken, which is the sum of the wind and the chicken. V_wind is coming from the east at 50m/s, so this vector can be expressed as V_wind = -50i. To find the vector representing the chicken's velocity respective to air, we find the unit vector of i+j: 1/sqrt(2)*(i+j) => sqrt(2)/2 * (i+j). Now, multiply this unit vector by the speed of the chicken. 100*(sqrt(2)/2 * (i+j)) => 50sqrt(2) * (i+j) = V_chicken.
Now we add the two, V_chicken + V_wind = 50*sqrt(2)i +50*sqrt(2)j - 50i = 20.710678i + 70.710678j = V_actualchicken. Find the magnitude of the velocity now with pythagorean theorem.. sqrt(20.71^2 + 70.71^2) = 73.68m/s. And then the direction, arctan(70.71/20.71) = 73.675 degrees N of E.
Thus we can find its position by doing 73.68m/s * 10 => 736.8m, which is the magnitude of displacement, and breaking it down into components.
736.8 cos 73.675 => 207.1i
736.8 sin 73.675 => 707.09j
Its position will be at (207.1, 707.09), why you ask? BECAUSE THE DUCK THAT INVITED TO CHICKEN TO THE BARBEQUE CLEVERLY SHOT THE CHICKEN DOWN A LITTLE BEFORE THAT POINT KNOWING THAT THE CHICKEN WOULD CERTAINLY FLEE LIKE A CHICKEN IF HE HAD KNOWN THE BARBEQUE'S MAIN DISH WAS THE CHICKEN AND HAD THE BARBEQUE THERE! HAHAHAHAHAHAHAHAHAHAHA, mm chicken (^)<

Problem 6: In a 100-m race, a duck and a chicken cross the finish line at the same time, taking 10.2s. Accelerating uniformly, the duck took 2s and the chicken took 3s to attain max speed, which they maintained for the rest of the race. a) What was the acceleration of the duck and of the chicken? b) What were their max speeds? c)Which sprinter was ahead at the 6s mark, and by how much?

a) We can divide up their trajectories into a part with acceleration and a part with constant velocity. d1 = v0t + 1/2a*t1^2 and then d2 = vt2. d = d1 + d2.
For the duck, d1 = 0 + .5*a*2^2. d2 = vt2, but from v = v0 + at, v is just a*t1, so d2 = a*t1*t2. We then have d = 2a + at1t2. From the givens, 10.2s = t1 + t2, and since t1 = 2s, we have t2 = 8.2s.

100m = 2a + a*2*8.2, solving for a... a = 5.435m/s^2 for the duck.
Great! Now we apply the same principle for the chicken, with t1 as 3 and t2 as 7.2 now. a for the chicken is then 3.831m/s^2.

b) To find the respective max speeds, we can use v = v0+at. Since they both started from rest, v0 = 0.
For the duck, we have v = 5.435*2 => 10.870m/s.
For the chicken, we have v = 3.831*3 => 11.493m/s.

c) For this part of the problem, we can use the equations for d1 and d2 obtained in part a.
For the duck, d1 = 0 + .5*5.435*2^2. then d2 is 10.870*4. The sum, d1 + d2, is then 54.35m.
For the chicken, d1 = 0 + .5*3.831*3^2 and d2 = 11.493*3. The displacement is then 51.72m.
The duck is ahead by 2.63m!

Problem 7: A chicken is dropped into a well. a) If the sound of the splash is heard 2.40 s later, how far below the top of the well is the surface of the water? The speed of sound in air for this case is 336m/s. b) If the travel time for the sound is neglected, how much percent error is introduced when the depth of the well is calculated?

a) The 2.40s given takes into account both the time it takes for the sound to travel back up to the observer and the time it takes for the chicken to hit the water. Since the sound in air travels the same distance as the chicken, we can set the displacement equations equal to each other. We'll want to use x-x0 = vt and x-x0 = v0 + 1/2at^2. The speed of sound in air is approximately constant, so we can just use 0 - h = v_snd*t_snd. For the chicken however, it is accelerating downwards at 9.8m/s^2. So we have h - 0 = 0 - 1/2*9.8*t_chk^2. Solving for h and setting both equations equal to each other, we have v_snd*t_snd = 4.9*t_chk^2. Since 2.40 = t_chk + t_snd, we can make a substitution. 336*t_snd = 4.9*(2.40-t_snd)^2. Solve for t_snd and get 0.0786s or 73.2928s. Take the first one since the second one doesn't make sense (the total time was 2.40s).
Now, since 0 - h = v_snd*t_snd, we can solve for h and find that the well is 26.41m deep.
b) If the travel time for the sound is neglected, then we can just do x-x0 = v0t + 1/2at^2 using t = 2.40s. x-x0 is then 28.22m. The percent error is the difference over the actual, so we have a % error of 6.85%.

Problem 8: A duck of 2.5kg. as the duck paddles a force of .1N acts on it in due east. the current of the water is .2N at 52deg south of east. When those forces begin to act, the velocity of the duck is .11 m/sec due east. Find the magnitude and direction of the displacement of the duck after 3 seconds.

F_x = 0.1*Cos + 0.2*Cos[(360-52)*Pi/180] = 0.223N
F_y = 0.1*Sin + 0.2*Sin[(360-52)*Pi/180] = -0.158N

a_x = F_x/m = 0.0893m/s^2
a_y = F_y/m = -0.0630m/s^2

x-x0 = v0t + 1/2at^2
x-x0 = 0.11*3 + .5*0.0893*3^2 = 0.73185m
y-y0 = 0 + 1/2at^2
y-y0 = 0 + .5*(-0.0630)*3^2 = -0.2835m

R = Sqrt[(y-y0)^2+(x-x0)^2]
R = Sqrt[(-0.2835)^2+(0.73185)^2] = 0.785m
Direction = ArcTan[(y-y0)/(x-x0)] = -0.37 rads = -21 degrees N of E