Kinematics and Calculus

Jerk is the derivative of acceleration. Undo that process. Integrate jerk to get acceleration as a function of time. I propose we call this the zeroth equation of motion for constant jerk. The reason why will be apparent after we finish the next derivation.

Acceleration is directly proportional to time when jerk is constant.

a ∝ t

Acceleration is the derivative of velocity. Integrate acceleration to get velocity as a function of time. We've done this process before. We called the result the velocity-time relationship or the first equation of motion when acceleration is constant. We should give it a similar name. This is the first equation of motion for constant jerk.

Velocity is proportional to time squared when jerk is constant.

v ∝ t²

Velocity is the derivative of displacement. Integrate velocity to get displacement as a function of time. We've done this before too. The resulting displacement-time relationship will be our second equation of motion for constant jerk.

Displacement is proportional to time cubed when jerk is constant.

s ∝ t³

Please notice something about these equations. When jerk is zero, they all revert back to the equations of motion for constant acceleration. Zero jerk means constant acceleration, so all is right with the equations we've created.

So where do we go next? Should we connect velocity-displacement (a third equation), acceleration-velocity (a fourth equation), acceleration-displacement (a fifth equation)? Look at that scary cubic equation for displacement. That can't be our friend.

The zeroth and third equations can be combined algebraically to make a fourth equation of motion for constant jerk. I won't show you any of the work — just the results, which should remind you of something. (It has the same form as the third equation of motion for constant acceleration.)

Acceleration is proportional to the square root of velocity when jerk is constant.

a ∝ √v

Algebra probably can't be used to generate the other two equations. Ask a mathematician if you need proof.

We can make it solvable if we just examine the special cases when s₀ = v₀ = a₀ = 0. Then…

Which means that…

…when jerk is constant.

We also know that when jerk is zero, we should at least get back what we started with (our initial values). What we don't know is what all the other stuff should be in between. We should get equations that look something like these…

This problem isn't done now, but I'm done with it for now.

where…

Draw it first.

1. Velocity is the first derivative of displacement.

   v =	ds	= 3t2 − 30t + 54
   	dt

2. Acceleration is the second derivative of displacement or the first derivative of velocity.

   a =	dv	= 6t − 30
   	dt

3. In general, the extreme values of a function occur at the endpoints of the domain or at local extrema (if they exist). Let's check the endpoints first. How fast is the object moving at the beginning and end of the time interval?

   v(0) =	3(0)2 − 30(0) + 54
   v(0) =	+54 m/s
   v(10) =	3(10)2 − 30(10) + 54
   v(10) =	+54 m/s

   Local extrema occur where the derivative of a function is zero. The extrema of the velocity function can be found at the places where the acceleration is zero.

   a =	6t − 30 = 0 m/s2
   t =	5 s

   This is the time where acceleration reversed direction. It corresponds to an inflection point on the graph.

   

   How fast was the object moving at this time?

   v(5) = 3(5)2 − 30(5) + 54 v(5) = −21 m/s

   We now have all the info needed to answer this question and the next one. The maximum velocity was +54 m/s. It occurred at 0 s and 10 s.

4. The minimum velocity was −21 m/s. It occurred at 5 s.

5. The direction of motion is determined by the sign of the velocity. Direction is reversed whenever the velocity switches sign. Between two signs lies a zero. Let's find the zeros.

   v =	3t2 − 30t + 54
   v =	3(t 2 − 10t + 18)

   t =	10 ± √[102 − 4(18)]
   	2

   t =	(5 ± √7) s
   t =	2.354… s, 7.645… s

   We have all the info we need to answer this question, but it's scattered. Let's make a little table summarizing what we know so far.

   event	time	velocity	direction
   start of problem	000.0 s	+54 m/s	forward
   direction reversal	002.3 s	+00 m/s	not moving
   minimum velocity	005.0 s	−21 m/s	backward
   direction reversal	007.6 s	+00 m/s	not moving
   end of problem	10.0 s	+54 m/s	forward

   The object was moving backward from 2.3 s to 7.6 s.

   

6. Set the position equation equal to zero to determine the times when the object was at the starting point (s = 0 m).

   s =	t3 − 15t2 + 54t	= 0 m
   s =	t(t − 6)(t − 9)	= 0 m
   t =	0 s, 6 s, 9 s

   

7. Average velocity is displacement divided by time. We know the object started at s = 0 m so that means ∆s = s(10 s)

   v =  ∆s ∆t

   v =  (10)3 − 15(10)2 + 54(10) 10

   v =  +4 m/s

8. Average speed is distance divided by time. Finding distance is not easy. We need to find the distance traveled during the three stages of motion — forward, backward, and forward again — and then add the absolute values of the displacements. Before we can do that, we need to actually find the critical positions.

   s =	t3 − 15t2 + 54t
   s(0) =	(0)3 − 15(0)2 + 54(0)
   s(0) =	0 m
   s(2.354) =	(2.354)3 − 15(2.354)2 + 54(2.354)
   s(2.354) =	+57.040… m
   s(7.645) =	(7.645)3 − 15(7.645)2 + 54(7.645)
   s(7.645) =	−17.040… m
   s(10) =	(10)3 − 15(10)2 + 54(10)
   s(10) =	+40 m

   Now combine them.

   segment	start	finish	distance
   moving forward	+00 m	+57 m	057 m
   moving backward	+57 m	−17 m	074 m
   moving forward	−17 m	+40 m	057 m
   overall			188 m

   Finally, divide distance by time.

   v =	∆s	=	188 m	= 18.8 m/s
   	∆t		10 s

1. A quick glance at the graph shows the elevator accelerating downward over 3 s, then coasting, then accelerating upward for 2 s, coasting again, and then accelerating for a brief burst before stopping. The speed increases, remains constant, decreases, remains constant, and decreases a bit more — all in the down direction. The greatest speed would happen at the end of the first triangular region on the acceleration-time graph. The area under this bit is the change in velocity from its initial value of zero. Therefore…

   ∆v = area under a-t graph ∆v = area of a triangle ∆v = ½bh ∆v = ½(3.0 s)(−6.0 m/s2) ∆v = −9.0 m/s

2. If an elevator is going to work properly it has to stop on a floor to let people off. Therefore, the velocity of the elevator at the end of the graph should be zero. Since change in velocity equal the area under the curve on an acceleration-time graph, we need the total area of the three triangular segments to add up to zero. We've already determined this change for the first triangular segment. Let's repeat it for the second.

   ∆v = area under a-t graph ∆v = area of a triangle ∆v = ½bh ∆v = ½(2.0 s)(+8.0 m/s2) ∆v = +8.0 m/s

   Add these two areas up and you don't get zero, you get…

   ∆v = −9.0 m/s + 8.0 m/s ∆v = −1.0 m/s

   Therefore, the area under the remaining segment must be +1.0 m/s to compensate. Use the same concepts as before, but this time solve for the base (change in time) of the triangle instead of the area (change in speed).

   ∆v =  ∆v =	area under a-t graph area of a triangle
   ∆v =  +1.0 m/s =	½bh ½∆t(+4.0 m/s2)
   t =	+0.50 s

3. The best way to construct the graphs for the next two questions is systematically — beginning from first principles. The rate of change of position is called velocity, the rate of change of velocity is called acceleration, and the rate of change of acceleration is called jerk. Yes, you heard me right — jerk. The straight line segments of the graph we started with correspond to intervals of constant jerk. (If they were curved we'd have non-uniform jerk.) Work backward, integrating the constant value of j to get a, then integrating that to get v, then integrating that to get s.

   j =	j	=							j
   a = ∫	j dt	=					a0	+	jt
   v = ∫	a dt	=			v0	+	a0t	+ ½	jt2
   s = ∫	v dt	=	s0	+	v0t	+ ½	a0t	+ ⅙	jt3

   Apply these equations over and over again. This much computation is best left to a computer. The results are summarized in the table below.

   A hydraulic elevator

   time (s)	jerk (m/s3)	acceleration (m/s2)	velocity (m/s)	position (m)
   0.00	0.0	0.0	0.00	0.000
   1.00	−0.4−	0.0	0.00	0.000
   2.50	0.4	−0.6−	−0.45−	−0.225−
   4.00	0.0	0.0	−0.90−	−1.350−
   13.00	0.8	0.0	−0.90−	−9.450−
   14.00	−0.8−	0.8	−0.50−	−10.217−
   15.00	0.0	0.0	−0.10−	−10.450−
   17.25	1.6	0.0	−0.10−	−10.675−
   17.50	−1.6−	0.4	−0.05−	−10.696−
   17.75	0.0	0.0	0.00	−10.700−
   20.00	0.0	0.0	0.00	−10.700−

4. Here's the velocity-time graph for the elevator. The horizontal segments correspond to intervals with no acceleration. The curved segments correspond to intervals with changing acceleration.

   

5. Here's the position-time graph for the elevator. The intervals with acceleration are curved. The intervals without acceleration are straight. The beginning and end of the graph are horizontal since the elevator is stopped.

   

6. The overall position of the elevator was 10.7 m below the second floor. Ceiling heights in a typical residential building are about 3 m. Public buildings like schools tend to have taller floors than homes. My guess is that the ceiling heights in this school are about 5 m. That would mean the elevator stopped in the basement.

Solutions…

1. Use the given equation to solve this part.

   v = a(1 − e^−t/b)

   Collect like terms.

   e^−t/b = 1 − v/a

   Undo the power of e with a natural logarithm.

   −t/b = ln(1 − v/a)

   Do a bit more algebra

   t = −b ln(1 − v/a)

   Numbers in.

   t = −(13.31 s) ln(1 − 111.111 m/s ÷ 128.1 m/s)

   Answer out.

   t = 26.89 s

2. There's an equation for average acceleration. Use it.

   a =	∆v
   	∆t

   The change in velocity was given and we just computed the time. Use these numbers

   a =	111.111 m/s
   	26.89 s

   Compute the answer.

   a = 4.132 m/s²

3. Let time approach infinity to determine the car's theoretical top speed. The limit of a negative exponent as it approaches infinity is zero.

   vmax =	lim ∆t→∞	a(1 − e−t/b)
   vmax = a(1 − 0) = a
   vmax = 128.1 m/s

4. Acceleration is the first derivative of velocity. Note that the a on the left is the quantity acceleration and the a on the right is a coefficient in the velocity-time function.

   a =  d  a(1 − e−t/b) dt

   a =  a  e−t/b b

5. The acceleration starts initially at t = 0 s with a high value.

   ai =	128.1 m/s	e−(0 s)/(13.31 s)
   	13.31 s
   a = 9.624 m/s2

6. The acceleration ends finally at t = 26.89 s with a low value.

   af =	128.1 m/s	e−(26.89 s)/(13.31 s)
   	13.31 s
   a = 1.276 m/s2

7. Displacement is the integral of velocity.

   ∆s =	t ⌠ ⌡ a(1 − e−t/b) dt 0
   ∆s =	t ⎡ ⎣ a(t + be−t/b) ⎤ ⎦ 0
   ∆s =	a(t + be−t/b) − a(0 + be−0/b)
   ∆s = a[t + b(e−t/b − 1)]

8. Since the car is only going forward, the distance traveled is the same as the magnitude of the displacement. Put numbers into the equation we just derived. Get an answer out.

   ∆s =	128.1 m/s[26.89 s + 13.31 s(e−26.89 s/13.31 s − 1)]
   ∆s =	1966 m

9. Use the basic definition of acceleration here. Since the acceleration is directed opposite the velocity (that is, the car is slowing down), the answer should have a negative sign.

   a =	∆v ∆t
   a =	−111.111 m/s
   	9.451 s
   a = −11.76 m/s2

10. List the givens and the unknown.

    v0 =	111.111 m/s
    v =	0 m/s
    t =	9.451 s
    a =	−11.76 m/s2
    ∆s =	?

    I can think of three ways to solve this if we assume the average acceleration is the acceleration — meaning the acceleration was constant.

    One — with the second equation of motion.

    ∆s =	v0t + ½at2
    ∆s =	(111.111 m/s)(9.451 s) + ½(−11.76 m/s2)(9.451 s)2
    ∆s =	525 m

    Two — with the third equation of motion.

    v2 =	v02 + 2a∆s
    ∆s =	−v02 2a
    ∆s =	−(111.111 m/s)2 2(−11.76 m/s2)
    ∆s =	525 m

    Three — with the mean speed rule.

    ∆s = vt ∆s = ½(v + v0)t ∆s = ½(0 m/s + 111.111 m/s)(9.451 s) ∆s = 525 m

11. Add the two partial distances to get the total distance from start to finish.

    ∆s = s1 + s2 ∆s = 1966 m + 525 m ∆s = 2491 m