Examples of Motion Problems In solving motion problems, you should always remember to start from the basics. For the discrete case, these are: v x-average =  x  t and a x-average =  v x  t.. For the continuous case when we have constant acceleration, these become: v(t) = v o + a*t and s(t) = s o + v o *t + ½ a*t 2 (where s can be either x or y depending on whether the motion is horizontal or vertical).

Example: Discrete Case Given the following values of velocity at the indicated times, find the average acceleration between t = 4 seconds and t = 5 seconds, and find the position at t = 5 seconds given that the initial position is -7 m. t (s) v (m/s) t (s) v (m/s)

Example: Discrete Case Diagram: t (s) v (m/s) t (s)v (m/s) x=0

Example: Discrete Case Since we are working with discrete values, we start with the definitions of average velocity and average acceleration: a x-average =  v x  t and v x-average =  x  t For the average acceleration between t = 4 s and t = 5 s, we can apply the average acceleration formula directly: a x-average =  v x  t becomes a avg =  v x  t = (7.95 m/s – 10 m/s) / (5s – 4s) = m/s 2.

Example: Discrete Case To find the position at t = 5 sec., we must start at the one position we know, which is x(t=0s) = x o = -7 m. and use the definition of average velocity: v x-average =  x  t, or  x = v x-average *  t or  x = x f – x i = v(between t f and t i )*(t f -t i ) However, because the velocity changes, we need to do this in individual steps:

Example: Discrete Case  x = x f – x i = v(between t f and t i )*(t f -t i ), or x f = x i + v(between t f and t i )*(t f -t i ) x 1 = x 0 + v(between t 1 and t 0 )*(t 1 -t 0 ) x 1 = -7 m + ½(-4.0 m/s m/s) * (1s – 0s) = -7 m + (-2.98 m/s)*(1 s) = m = x 1 x 2 = x 1 + v(between t 2 and t 1 )*(t 2 -t 1 ) x 2 = m + ½(-1.95 m/s m/s) * (2s – 1s) = m + (.53 m/s)*(1 s) = m = x 2 x 3 = m + ½(3.0 m/s m/s) * (3s – 2s) = m + (5.48 m) = m = x 3

Example: Discrete Case x 4 = m + ½(7.95 m/s m/s) * (4s – 3s) = m + (+8.98 m/s)*(1 s) = 5.00 m = x 4 x 5 = 5.00 m + ½(10.0 m/s m/s) * (5s – 4s) = 5.00 m + (+8.98 m) = m = x 5 In this particular case, the velocity data was obtained from the functional form: v(t) = +3 m/s – (7 m/s)*cos({  rad/4 sec}  t) [note that {  rad / 4 sec} = {45 o /sec} ] Which leads (via calculus) to functional forms for x(t) and a(t) of: a(t) = (7*  m/ 4 s 2 )*sin({  rad/4 sec}  t) x(t) = -7 m + (3 m/s)*t – (28 m/  sin({  rad/4 sec}  t)

Example: Discrete Case a(t) = (7*  m/ 4 s 2 )*sin({  rad/4 sec}  t) x(t) = -7 m + (3 m/s)*t – (28 m/  sin({  rad/4 sec}  t) If we use the above functional forms to calculate the acceleration between t = 4 sec and t = 5 sec, that is, at t = 4.5 sec, we get: a(t) = (7*  m/ 4 s 2 )*sin({  rad/4 sec}  4.5 sec) = m/s 2 which compares to our numerical result of m/s 2. If we use the above functional forms to calculate the position at t = 5 sec, we get: x(5 sec) = -7 m + (3 m/s)*(5 s) – (28 m/  sin({  rad/4 sec}  (5 s)) = -7 m + 15 m m = m which compares to our numerical result of m.

Example: Discrete Case a (functional result) = m/s 2 a (numerical result) = m/s 2 x (functional result) = m x (numerical result) = m Note that the functional and numerical results are close but not exactly the same. The numerical result is based on discrete data which is incomplete, whereas the functional result is based on complete (continuous) data.

Example: plane taking off A plane takes off from an aircraft carrier. Assume the following: the plane accelerates with a constant acceleration, the flight deck is 80 meters long, the initial speed of the plane is zero, the final speed of the plane is 85 m/s. What is the acceleration of the plane during take- off, and how long a time does the take-off take?

Plane taking off – continued Draw a diagram:  x = 80 meters v o = 0 m/sv f = 85 m/s t o = 0 s t f = ? a = ?

Plane taking off – continued Since this is a constant acceleration problem along the horizontal, we have two equations: v(t) = v o + a*t and x(t) = x o + v o *t + ½ a*t 2. We are given the following: v o = 0 m/s; v f = 85 m/s; x f – x o = 80 meters (from this we can let x o = 0 m, so x f = 80 m). We are asked for the acceleration, a, and the time, t.

Plane taking off – continued Since we have two equations and two unknowns (a and t), we should be able to solve for both a and t. We now substitute the knowns into our two equations: v(t) = v o + a*t becomes 85 m/s = 0 m/s + a*t and x(t) = x o + v o *t + ½ a*t 2 becomes 80 m = 0 m + (0 m/s)*t + ½ a*t 2.

Plane taking off – continued 85 m/s = 0 m/s + a*t 80 m = 0 m + (0 m/s)*t + ½ a*t 2 Note that both equations have both unknowns, so we have to use the techniques of simultaneous equations. If we solve for a in the first equation, we get: a = 85 m/s / t, and now we use this in the second equation: 80 m = 0 m + (0 m/s)*t + ½ (85 m/s / t)*t 2.

Plane taking off – continued 80 m = 0 m + (0 m/s)*t + ½ (85 m/s / t)*t 2 This is now one equation in one unknown (t) 80 m = (42.5 m/s)*t, or t = (80 m) / (42.5 m/s) = 1.88 seconds. We can now use this value of t in our first equation: 85 m/s = 0 m/s + a*t to get: 85 m/s = 0 m/s + a*(1.88 sec), or a = (85 m/s) / (1.88 sec) = m/s 2.

Plane taking off – continued t = 1.88 seconds. a = m/s 2. Are these results reasonable? A plane taking off from an aircraft carrier on a short flight deck certainly has to take off in a short time, so about 2 seconds looks reasonable. The acceleration of m/s 2 is equivalent to 4.61 gees’ (1 gee = 9.8 m/s 2 ). This is a pretty big acceleration, but that also sounds reasonable given the circumstances. By the way, this type of acceleration will tend to make most people black out – not a good thing when taking off in a plane.

Example: Throwing an object up A candy bar is thrown upwards from a person on the ground to a person on a balcony 15 meters above the ground. The person on the ground throws the candy bar up at a speed of 3 m/s. How long will it take the candy bar to reach the person on the balcony?

Example: Throwing an object up We recognize this as a “falling” problem, so we can use the equations for constant acceleration with the acceleration equaling that of gravity. y o = 0 m v o = 3 m/s t o = 0 s y = 15 m v = ? t = ? a = g = -9.8 m/s 2

Example: Throwing an object up v(t) = v o + a*t and y(t) = y o + v o *t + ½ a*t 2 where we identify the following: y = 15 m (final position is 15 meters above the ground) y o = 0 m (initial position is on the ground) v = ? (speed when it reaches the person on the balcony) v o = 3 m/s (initial speed ) t = ? (time when it reaches the person on the balcony) a = g = -9.8 m/s 2 (acceleration due to gravity)

Example: Throwing an object up v(t) = v o + a*t and y(t) = y o + v o *t + ½ a*t 2 Substituting in the values we know: v(t) = (3 m/s) + (-9.8 m/s 2 ) *t and 15 m = 0 m + (3 m/s)*t + ½ (-9.8 m/s 2 )*t 2 We see that we have two equations for two unknowns, but the second equation has only one unknown, t. However, it is a quadratic equation in that unknown.

Example: Throwing an object up v(t) = (3 m/s) + (-9.8 m/s 2 ) *t and 15 m = 0 m + (3 m/s)*t + ½ (-9.8 m/s 2 )*t 2 To solve quadratic equations, we need to put this in standard form, a*x 2 + b*x + c = 0 ½ (-9.8 m/s 2 )*t 2 + (3 m/s)*t - 15 m = 0 m and then use the quadratic formula x = [-b ± {b 2 – 4*a*c} 1/2 ] / [2*a] Where a = -4.9 m/s 2, b = 3 m/s, and c = -15 m.

Example: Throwing an object up x = [-b ± {b 2 – 4*a*c} 1/2 ] / [2*a] where a = -4.9 m/s 2, b = 3 m/s, and c = -15 m. t = [-(3 m/s) ± {(3 m/s) 2 – 4*(-4.9 m/s 2 )*(-15 m)} 1/2 ] / [2*(-4.9 m/s 2 )] = [-(3 m/s) ± {-285 m 2 /s 2 } 1/2 ] / [-9.8 m/s 2 ]. Note that we have the square root of a negative number in the formula. This means that the math gives us an imaginary number for the time. What does this mean physically?

Example: Throwing an object up t = [-(3 m/s) ± {-285 m 2 /s 2 } 1/2 ] / [-9.8 m/s 2 ]. What does this mean physically? You may have noticed that the person on the ground did not throw the candy bar up with a very big speed, only 3 m/s. What happens is that the candy bar never makes it up to 15 meters. We only “imagine” that the candy bar gets up that high. The math is trying to tell us that the candy bar never reaches that high. So what do we have to do to get the candy bar that high?

Example: Throwing an object up x = [-b ± {b 2 – 4*a*c} 1/2 ] / [2*a] where a = -4.9 m/s 2, b = 3 m/s, and c = -15 m. We see from the formula that to make x real, the quantity b 2 has to be greater than (4*a*c). In our case, to make t real, the initial velocity (which is the “b”) must be at least as big as the quantity [ 4*(½g)*(y)] (which is the “4*a*c”). Therefore, let’s retry this problem with a bigger initial speed, say 20 m/s.

Example: Throwing an object up x = [-b ± {b 2 – 4*a*c} 1/2 ] / [2*a] where a = -4.9 m/s 2, b = 20 m/s, and c = -15 m. t = [-(20 m/s) ± {(20 m/s) 2 – 4*(-4.9 m/s 2 )*(-15 m)} 1/2 ] / [2*(-4.9 m/s 2 )] = [-(20 m/s) ± {106 m 2 /s 2 } 1/2 ] / [-9.8 m/s 2 ]. What does the plus or minus sign in the formula mean? It actually gives us two answers: t = 0.99 seconds, and t = 3.09 seconds. If we look at our diagram, we see that the candy bar passes the person on the way up (at t = 0.99 s) and on the way back down (at t = 3.09 s).

Example: Throwing an object up Now that we know the times when the candy bar reaches the person on the balcony, we could use the other (velocity) equation to find the speeds at these two times. v(t) = (20 m/s) + (-9.8 m/s 2 ) *t On the way up, t = 0.99 s, so v(up) = (20 m/s) + (-9.8 m/s 2 ) *(0.99 s) = m/s, and on the way down, t = 3.09 s, so v(down) = (20 m/s) + (-9.8 m/s 2 ) *(3.09 s) = m/s.