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A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship’s acceleration look like?

Line graph of velocity versus time. The line has three legs. The first leg is flat. The second leg has a negative slope. The third leg also has a negative slope, but the slope is not as negative as the second leg.

(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.

(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.

A line graph of acceleration versus time. There are three legs of the graph. All three legs are flat and straight. The first leg shows constant acceleration of 0. The second leg shows a constant negative acceleration. The third leg shows a constant negative acceleration that is not as negative as the second leg.
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Section summary

  • Graphs of motion can be used to analyze motion.
  • Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
  • The slope of a graph of displacement x size 12{x} {} vs. time t size 12{t} {} is velocity v size 12{v} {} .
  • The slope of a graph of velocity v size 12{v} {} vs. time t size 12{t} {} graph is acceleration a size 12{a} {} .
  • Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.

Conceptual questions

(a) Explain how you can use the graph of position versus time in [link] to describe the change in velocity over time. Identify (b) the time ( t a , t b , t c , t d , or t e ) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative.

Line graph of position versus time with 5 points labeled: a, b, c, d, and e. The slope of the line changes. It begins with a positive slope that decreases over time until around point d, where it is flat. It then has a slightly negative slope.
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(a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in [link] . (b) Identify the time or times ( t a , t b , t c , etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At which times is it negative?

Line graph of position over time with 12 points labeled a through l. Line has a negative slope from a to c, where it turns and has a positive slope till point e. It turns again and has a negative slope till point g. The slope then increases again till l, where it flattens out.
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(a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in [link] . (b) Based on the graph, how does acceleration change over time?

Line graph of velocity over time with two points labeled. Point P is at v 1 t 1. Point Q is at v 2 t 2. The line has a positive slope that increases over time.
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(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in [link] . (b) Identify the time or times ( t a , t b , t c , etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative?

Line graph of velocity over time with 12 points labeled a through l. The line has a positive slope from a at the origin to d where it slopes downward to e, and then back upward to h. It then slopes back down to point l at v equals 0.

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Consider the velocity vs. time graph of a person in an elevator shown in [link] . Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip.

Line graph of velocity versus time. Line begins at the origin and has a positive slope until it reaches 3 meters per second at 3 seconds. The slope is then zero until 18 seconds, where it becomes negative until the line reaches a velocity of 0 at 23 seconds.
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A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.

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Problems&Exercises

Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.

(a) By taking the slope of the curve in [link] , verify that the velocity of the jet car is 115 m/s at t = 20 s size 12{t="20"`s} {} . (b) By taking the slope of the curve at any point in [link] , verify that the jet car’s acceleration is 5 . 0 m/s 2 size 12{5 "." "0 m/s" rSup { size 8{2} } } {} .

Line graph of position over time. Line has positive slope that increases over time.
Line graph of velocity versus time. Line is straight with a positive slope.

(a) 115 m/s size 12{"115 m/s"} {}

(b) 5 . 0 m/s 2 size 12{5 "." "0 m/s" rSup { size 8{2} } } {}

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Using approximate values, calculate the slope of the curve in [link] to verify that the velocity at t = 10.0 s size 12{t="10"`s} {} is 0.208 m/s. Assume all values are known to 3 significant figures.

Line graph of position versus time. Line is straight with a positive slope.
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Using approximate values, calculate the slope of the curve in [link] to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

v = ( 11.7 6.95 ) × 10 3 m ( 40 . 0 – 20 .0 ) s = 238 m/s
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By taking the slope of the curve in [link] , verify that the acceleration is 3 . 2 m /s 2 at t = 10 s size 12{t="10"`s} {} .

Line graph of velocity versus time. Line has a positive slope that decreases over time until the line flattens out.
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Construct the displacement graph for the subway shuttle train as shown in [link] (a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

Line graph of position versus time. Line begins with a slight positive slope. It then kinks to a much greater positive slope.
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(a) Take the slope of the curve in [link] to find the jogger’s velocity at t = 2 . 5 s size 12{t=2 "." 5`s} {} . (b) Repeat at 7.5 s. These values must be consistent with the graph in [link] .

Line graph of position over time. Line begins sloping upward, then kinks back down, then kinks back upward again.
Line graph of velocity over time. Line begins with a positive slope, then kinks downward with a negative slope, then kinks back upward again. It kinks back down again slightly, then back up again, and ends with a slightly less positive slope.
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A graph of v t is shown for a world-class track sprinter in a 100-m race. (See [link] ). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t = 5 s ? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

Line graph of velocity versus time. The line has two legs. The first has a constant positive slope. The second is flat, with a slope of 0.

(a) 6 m/s

(b) 12 m/s

(c) 3 m/s 2 size 12{"3 m/s" rSup { size 8{2} } } {}

(d) 10 s

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[link] shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.

Line graph of position versus time. The line has 4 legs. The first leg has a positive slope. The second leg has a negative slope. The third has a slope of 0. The fourth has a positive slope.
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Questions & Answers

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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there is enough information to calculate an AVERAGE acceleration
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mistake, there is enough information to calculate an average velocity
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It is the opposite of kinetic friction
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static fiction is friction between two surfaces in contact an none of sliding over on another, while Kinetic friction is friction between sliding surfaces in contact.
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I don't get it,if it's static then there will be no friction.
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It means that static friction is that friction that most be overcome before a body can move
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static friction is a force that keeps an object from moving, and it's the opposite of kinetic friction.
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It is a force a body must overcome in order for the body to move.
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Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration. To change from weight to mass divide by gravity (9.8 m/s^2).
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the write question should be " How many Topics are in O- Level Physics, or other branches of physics.
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Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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