motion 1D notes

1. Upward motion is defined as a positive velocity and downward velocity is defined as a negative velocity

2. If the motion is horizontal, going right is a positive velocity and going left is a negative velocity.

3. Speed, on the other hand, is always positive (or zero). If a car goes by at 50 mph, for instance, you say its speed is 50, and you mean positive 50, regardless of whether it's going to the right or the left.

4. If an object is going down (or to the left) faster and faster, its speed is increasing, but its velocity is decreasing because its velocity is becoming a bigger negative (and bigger negatives are smaller numbers).

5. Acceleration is defined as the rate of change of velocity, not speed.6. If an object is slowing down while going in the downward direction, and thus has an

increasing velocity — because the velocity is becoming a smaller negative — the object has a positive acceleration.

7. An object is speeding up (what we call “acceleration” in everyday speech) whenever the velocity and the calculus acceleration are both positive or both negative.

8. An object is slowing down (what we call “deceleration”) when the velocity and the calculus acceleration are of opposite signs.

9. Maximum and minimum height of H(t) occur at the local extrema. To locate them, set the derivative of H(t) — that’s V(t) — equal to zero and solve.

10. These two numbers are the zeros of V(t) and the t-coordinates — that’s time-coordinates — of the max and min of H(t)

11. Total displacement is defined as the final position minus the initial position.12. Average velocity is given by total displacement divided by elapsed time.13. Maximum and minimum velocity during the interval from t1to t2 seconds are determined

with the derivative of V(t): Set the derivative of V(t) — that’s A(t) — equal to zero and solve for t. Calculate V(t), V(t1) and V(t2).

Positive velocity

Negative velocity Positive velocity

Negative velocity

14. Finding direction change: Set velocity function V(t) =0, and solve for t, when v=0. Check to see if there is change of sign at time, because if something stops, it does not mean change of direction.(eg Car at red light)

15. Total distance travelled is determined by adding up the distances travelled on each lega. If particle does not change direction distance = |Xf – Xi|b. If particle chnges direction at Xc distance = |Xf – Xc| + |Xc – Xi|

16. Average speed is given by the total distance travelled divided by the elapsed time.17. Minimum speed: For a continuous velocity function, the minimum speed is zero whenever

the maximum and minimum velocities are of opposite signs or when one of them is zero. When the maximum and minimum velocities are both positive or both negative, then the minimum speed is the lesser of the absolute values of the maximum and minimum velocities.

18. Maximum speed: In all cases, the maximum speed is the greater of the absolute values of the maximum and minimum velocities.

19. Maximum and minimum acceleration: To find the acceleration’s min and max from t1 to t2, set the derivative of A(t) equal to zero and solve by A’(t) = 0. If this equation has no solutions, there are no critical numbers and thus the absolute extrema must occur at the interval’s endpoints.

20. Note that when the acceleration is negative that means that the velocity is decreasing. When the acceleration is positive the velocity is increasing.

21. A negative acceleration can mean…(a) An object moving in the positive direction (positive velocity) is slowing down.. or (b) An object moving in the negative direction (negative velocity) is speeding up.

22. When the directions (signs) of velocity and acceleration are the same (positive or negative), the object is speeding up.

23. When velocity and acceleration have opposite directions (signs), the object slows down. 24. how do you keep track of all these types of acceleration. “Speeding up backwards

means… I don’t know!”

Try this. We usually visualize speeding up as positive, and slowing down as negative. I put

that in the top row. A positive velocity means it’s going in the positive direction (like forwards), and a

negative direction is backwards.

Now play a little “Battleship.” o positive X positive = positiveo positive X negative = negativeo negative X negative = positive

So, reading off the chart, an object moving backwards (-v) that is going faster and faster (speeding up) means it has a negative acceleration.

25. Gradient (Slope or Tan ) of distance-time graph = speed26. Gradient (Slope or Tan ) of displacement-time graph = velocity27. Area under a velocity-time graph gives the displacement.28. Area under a speed-time graph gives the distance.29. Gradient (Slope or Tan ) of velocity-time graph = Acceleration30. Negative slope ∴Negative acceleration, (or deceleration. Car is slowing down)31.

Graph type x - t v - t 1. a - t

Gradient Velocity Acceleration MeaninglessArea under graph Meaningless Displacement Change in velocity

32. Calculus of 1D

v=dxdt

a=dvdt

a=(dvdx )( dx

dt )=v( dvdx )

vdv=a ( x )d x

∫ v dv=∫a ( x ) dx

v2

2=∫ a ( x ) dx (when acceleration is given as function

of x)

v (x)=dxdt

dxv ( x )

=d t

∫ dxv ( x )

=∫ dt

∫ dxv ( x )

=t (when velocity is given as

function of x)

33.

How to Analyze Position, Velocity, and Acceleration with Differentiation Every time you get in your car, you witness differentiation first hand. Your speed is the first derivative of your position. And when you step on the accelerator or the brake — accelerating or decelerating — you experience a second derivative.

If a function gives the position of something as a function of time, the first derivative gives its velocity, and the second derivative gives its acceleration. So, you differentiate position to get velocity, and you differentiate velocity to get acceleration.

Here’s an example. A yo-yo moves straight up and down. Its height above the ground, as a function of time, is given by the function

where t is in seconds and H(t) is in inches. At t = 0, it’s 30 inches above the ground, and after 4 seconds, it’s at height of 18 inches.

The yo-yo’s height, from 0 to 4 seconds.

Velocity,V(t) is the derivative of position (height, in this problem), and acceleration, A(t), is the derivative of velocity. Thus

The graphs of the yo-yo’s height, velocity, and acceleration functions from 0 to 4 seconds.

Velocity versus speed. Your friends won’t complain — or even notice — if you use the words “velocity” and “speed” interchangeably, but your friendly mathematician will complain. Here’s the difference.

For the velocity function in the figure above, upwardmotion is defined as a positive velocity and downward velocity is defined as a negative velocity — this is the standard way velocity s treated in most calculus and physics problems. (If the motion is horizontal, going right is a positive velocity and going left is a negative velocity.)

Speed, on the other hand, is always positive (or zero). If a car goes by at 50 mph, for instance, you say its speed is 50, and you mean positive 50, regardless of whether it's going to the right or the left. For velocity, the direction matters; for speed it doesn’t. In everyday life, speed is a simpler idea than velocity because it agrees with common sense. But in calculus, speed is actually the trickier idea because it doesn’t fit nicely in the three-function scheme shown in the figure above.

You’ve got to keep the velocity-speed distinction in mind when analyzing velocity and acceleration. For example, if an object is going down (or to the left) faster and faster, its speed is increasing, but its velocity is decreasing because its velocity is becoming a bigger negative (and bigger negatives are smaller numbers). This seems weird, but that’s the way it works. And here’s another strange thing: Acceleration is defined as the rate of change of velocity, not speed. So, if an object is slowing down while going in the downward direction, and thus has an increasing velocity — because the velocity is becoming a smaller negative — the object has a positive acceleration. In everyday English, you’d say the object is decelerating (slowing down), but in calculus class, you say that the object has a negative velocity and a positive acceleration. (By the way, “deceleration” isn’t exactly a technical term, so you should probably avoid it in calculus class. It’s best to use the following vocabulary: “positive acceleration,” “negative acceleration,” “speeding up,” and “slowing down.”

Maximum and minimum heightof H(t) occur at the local extrema you see in the above figure. To locate them, set the derivative of H(t) — that’s V(t) — equal to zero and solve.

These two numbers are the zeros of V(t) and the t-coordinates — that’s time-coordinates — of the max and min of H(t), which you can see in the second figure. In other words, these are the times when the yo-yo reaches its maximum and minimum heights. Plug these numbers into H(t) to obtain the heights:

H (0.47) ≈ 31.1

H (3.53) ≈ 16.9

So, the yo-yo gets as high as about 31.1 inches above the ground at t ≈ 0.47 seconds and as low as about 16.9 inches at t ≈ 3.53 seconds.

Total displacement is defined as the final position minus the initial position. So, because the yo-yo starts at a height of 30 and ends at a height of 18,

Total displacement = 18 – 30 = –12.

This is negative because the net movement is downward.

Average velocityis given by total displacement divided by elapsed time. Thus,

This negative answer tells you that the yo-yo is, on average, going down 3 inches per second.

Maximum and minimum velocityof the yo-yo during the interval from 0 to 4 seconds are determined with the derivative of V(t): Set the derivative of V(t) — that’s A(t) — equal to zero and solve:

Now, evaluate V(t) at the critical number, 2, and at the interval’s endpoints, 0 and 4:

So, the yo-yo has a maximum velocity of 5 inches per second twice — at both the beginning and the end of the interval. It reaches a minimum velocity of –7 inches per second at t = 2 seconds.

Total distance traveledis determined by adding up the distances traveled on each leg of the yo-yo’s trip: the up leg, the down leg, and the second up leg.

First, the yo-yo goes up from a height of 30 inches to about 31.1 inches (where the first turn-around point is). That’s a distance of about 1.1 inches. Next, it goes down from about 31.1 to about 16.9 (the height of the second turn-around point). That’s a distance of 31.1 minus 16.9, or about 14.2 inches. Finally, the yo-yo goes up again from about 16.9 inches to its final height of 18 inches. That’s another 1.1 inches. Add these three distances to obtain the total distance traveled: ~1.1 + ~14.2 + ~1.1 ≈ 16.4 inches.

Average speedis given by the total distance traveled divided by the elapsed time. Thus,

Maximum and minimum speed. You previously determined the yo-yo’s maximum velocity (5 inches per second) and its minimum velocity (-7 inches per second). A velocity of -7 is a speed of 7, so that’s the yo-yo’s maximum speed.Its minimum speed of zero occurs at the two turnaround points.

For a continuous velocity function, the minimum speed is zero whenever the maximum and minimum velocities are of opposite signs or when one of them is zero. When the maximum and minimum velocities are both positive or both negative, then the minimum speed is the

lesser of the absolute values of the maximum and minimum velocities. In all cases, the maximum speed is the greaterof the absolute values of the maximum and minimum velocities. Is that a mouthful or what?

Maximum and minimum acceleration may seem pointless when you can just look at the graph of A(t) and see that the minimum acceleration of –12 occurs at the far left when t = 0 and that the acceleration then goes up to its maximum of 12 at the far right when t = 4. But it’s not inconceivable that you’ll get one of those incredibly demanding calculus teachers who has the nerve to require that you actually do the math and show your work — so bite the bullet and do it.

To find the acceleration’s min and max from t = 0 to t = 4, set the derivative of A(t) equal to zero and solve:

This equation, of course, has no solutions, so there are no critical numbers and thus the absolute extrema must occur at the interval’s endpoints, 0 and 4.

You arrive at the answers you already knew.

Note that when the acceleration is negative — on the interval [0, 2) — that means that the velocity is decreasing. When the acceleration is positive — on the interval (2, 4] — the velocity is increasing.

Speeding up and slowing down. Figuring out when the yo-yo is speeding up and slowing down is probably more interesting and descriptive of its motion than the info above. An object is speeding up (what we call “acceleration” in everyday speech) whenever the velocity and the calculus acceleration are both positive or both negative. And an object is slowing down (what we call “deceleration”) when the velocity and the calculus acceleration are of opposite signs.

Look at all three graphs in the figure above again. From t = 0 to about t = 0.47 (when the velocity is zero), the velocity is positive and the acceleration is negative, so the yo-yo is

slowing town (until it reaches its maximum height). When t = 0, the deceleration is greatest (12 inches per second per second; the graph shows an acceleration of negative 12, but here we’re calling it a deceleration so the 12 is positive). From about t = 0.47 to t = 2, both velocity and acceleration are negative, so the yo-yo is slowing down again (until it bottoms out at the lowest height). Finally, from about t = 3.53 to t = 4, both velocity and acceleration are positive, so the yo-yo is speeding up again. The yo-yo reaches its greatest acceleration of 12 inches per second per second at t = 4 seconds.

Tying it all together. Note the following connections among the three graphs in the figure above. The negative section of the graph of A(t) – from t = 0 to t = 2 – corresponds to a decreasing section of the graph of V(t) and a concave down section of the graph H(t). The positive interval of the graph of A(t)– from t = 2 to t = 4 – corresponds to an increasing interval on the graph of V(t) and a concave up interval on the graph H(t). When t = 2 seconds, A(t) has a zero, V(t) has a local minimum, and H(t) has an inflection point.

How Positive and Negative Acceleration Relate to Speed and Velocity In physics, the sign of an object’s acceleration depends on its direction. If you slow down to a complete stop in a car, for example, and your original velocity was positive and your final velocity was 0, so your acceleration is negative because a positive velocity came down to 0. However, if you slow down to a complete stop in a car and your original velocity was negative and your final velocity was 0, then your acceleration would be positive because a negative velocity increased to 0.

When you hear that acceleration is going on in an everyday setting, you typically think that means the speed is increasing. However, in physics, that isn’t always the case. An acceleration can cause speed to increase, decrease, and even stay the same!

Acceleration tells you the rate at which the velocity is changing. Because the velocity is a vector, you have to consider the changes to its magnitude and direction. The acceleration can change the magnitude and/or the direction of the velocity. Speed is only the magnitude of the velocity.

Here’s a simple example that shows how a simple constant acceleration can cause the speed to increase and decrease in the course of an object’s motion. Say you take a ball, throw it straight up in the air, and then catch it again. If you throw the ball upward with a speed of 9.8 m/s, the velocity has a magnitude of 9.8 m/s in the upward direction. Now the ball is under the influence of gravity, which, on the surface of the Earth, causes all free-falling objects to undergo a vertical acceleration of –9.8 m/s2. This acceleration is negative because its direction is vertically downward.

With this acceleration, what’s the velocity of the ball after 1.0 second? Well, you know that

Rearrange this equation and plug in the numbers, and you find that the final velocity after 1.0 second is 0 meters/second:

After 1.0 second, the ball has zero velocity because it’s reached the top of its trajectory, just at the point where it’s about to fall back down again. So the acceleration has actually slowed down the ball because it was going in the direction opposite the velocity.

Now see what happens as the ball falls back down to Earth. The ball has zero velocity, but the acceleration due to gravity accelerates the ball downward at a rate of –9.8 m/s2. As the ball falls, it gathers speed before you catch it. What’s its final velocity as you catch it, given that its initial velocity at the top of its trajectory is zero?

The time for the ball to fall back down to you is just the same as the time it took to reach the top of its trajectory, which is 1.0 second, so you can find the final velocity for this part of the ball’s motion with this calculation:

So the final velocity is 9.8 meters/second directed straight downward. The magnitude of this velocity — that is, the speed of the ball — is 9.8 meters/second. The acceleration increases the speed of the ball as it falls because the acceleration is in the same direction as the velocity for this part of the ball’s trajectory.

When you work with physics problems, bear in mind that acceleration can speed up or slow down an object, depending on the direction of the acceleration and the velocity of the object. Don’t simply assume that just because something is accelerating its speed must be increasing.

1. An object moves around the outside of a circle and ends up back where it began.

The object's displacement is1. Positive2. Negative3. Zero

2. An object moves around the outside of a circle and ends up back where it began.

The object's velocity is1. Positive2. Negative3. Zero

3. An object moves from +20 m away to +5 m away.


4. An object moves to the left.


5. An object going -24 m/s ends up going -12 m/s.

The object's acceleration is1. Positive2. Negative3. Zero

6. An object moves from -5 m away to 3 m away.


7. An object moves to the right.




9. A stopped object ends up going -8 m/s.


10. An object moves from -3 m away to -12 m away.


11. A stopped object ends up going 10 m/s.


12. An object +4m away moves to +8m away.




14. An object moves from -3 m away to -12 m away.


15. An object moves from +20 m away to +5 m away.


16. An object going 6 m/s stops.


17. An object going 3 m/s ends up going 10 m/s.


18. An object going -2 m/s ends up going -16 m/s.


19. An object moves to the left.




21. An object moves to the right.




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motion 1D notes