Understanding Simple Kinematics: Graphing Speed and Distance Traveled

  • Thread starter liamporter1702
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In summary, the conversation discusses a problem with finding the total distance traveled by a car based on a graph of its velocity over time. The original poster is having trouble with the question and is unsure where they went wrong in their attempt at a solution. They share their graph and method of finding the area under the graph, but another user points out that they have lumped together different sections of time and must use different acceleration values. The conversation then discusses a more efficient equation for finding the distance traveled and how to use it in this scenario.
  • #1
liamporter1702
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Hi guys, having some trouble with this question as I cannot see where I am going wrong. Very simple question but I must be misunderstanding something.

Homework Statement


The following table shows how the velocity of a car varied with time:

Velocity (km/h) 0 19 35 48 48 27 0
Time (s) 0 3 6 9 12 15 18

Assuming these speed values to be joined by straight lines, plot a graph of speed against time and determine the total distance travelled.

Homework Equations


s = dxt
Area under graph = distance travelled

The Attempt at a Solution


I assumed this question would be as simple as finding the area under the graph to get the total distance travelled. So I went about drawing my graph along with converting all the values for speeds into (m/s). I have attached a picture of what my graph looks like in case anyone would like to see it (its a bit messy). I'm sure there is a very simple explanation to where I am going wrong but I'm just not seeing it myself.

Thanks in advanced guys!
 

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  • #2
So what did you get and why do you think it is wrong ?
 
  • #3
Sorry forgot to put that info in my original post. I keep getting 139.97m by using the area under the graph rule. I do (13.33x9)/2 + (13.33x3) + (13.33x6)/2 = 139.97. Yet my questions sheet gives me an answer of 147m.
 
  • #4
Problem I have with your plot is not that it's messy (it really isn't). I have the same numbers but my plot looks quite diferent!

Problem I have is that you use different step sizes on the time axis! the 3 sec from 9-12 looks a lot longer than from 12-15.

This also suggests a linear constant acceleration from 0-9, which is not correct (otherwise the speeds would look x, 2x, 3x for t = 0,6,9 and they don't).

What do you do to determine the area under the graph ? 9 x 13.3 /2 + 3 x 13.3 + 6 x 13.3 /2 ?

[edit] well this guess of mine crossed your addendum.

Draw it with equidistant time intervals and you'll be OK.
 
  • #5
My plot was just a quick sketch as I was unsure how the graph would look, would you recommend drawing it out properly on graph paper?

I used that to get that area under graph based on splitting the area into two triangles and a rectangle. I then used 1/2 x base x height to get the area of the two triangles and added these onto the area of the rectangle.
 
  • #6
Either that, or you go to ##s=s_0 + v_0 \ t + {1\over 2} a \ t^2##, where ## a_{\ t-3\ \rightarrow\ t} = {v(t) - v(t-3) \over t - (t-3)} ## in steps of 3 sec

But: did you pick up that 0-9 sec cannot be lumped together because the accelerations in these three sections are not the same ?
Idem 12-18 sec.
 
  • #7
I think that is where I have gone wrong yes. I have lumped together 0-9 seconds, thank you for pointing that out! I am currently working through it again but this time I have drawn the graph out properly and I am splitting the area beneath the graph into triangles, trapeziums and a rectangle, then adding all these areas together, rather than lumping parts together. Fingers crossed it works this time!
 
  • #8
Let us know if you don't get 147.5 m !

By the way, do you realize you are doing exactly the same thing as this s = s0 + v0 t + 1/2 a t^2 , when you do it by adding triangles and rectangles ?
 
  • #9
I did get it this time! But if that equation is a faster way to get the answer then I'll try that also. What is that equation? Is it a SUVAT equation because I don't think I've seen that one. what does s0 and v0 represent?
 
  • #10
s0 and v0 are distance and speed at start.
So first interval: 0 and 0. acceleration a is (5.28 - 0 )/3 = 1.76 m/s2. s = 0 + 0 x 3 + 1/2 x 1.76 x 32 = 7.92 m .
Then second interval: 7.92 m and 5.28 m/s , a = (9.72-5.28)/3 = 1.48 m/s2.
s = 7.92 m + 5.28 * 3 + 1/2 * 1.48 * 32 = 30.42 m
the 5.28 * 3 is a rectangle area and the 1/2 * 1.48 * 32 is a triangle area.
etc etc
 

FAQ: Understanding Simple Kinematics: Graphing Speed and Distance Traveled

1. What is kinematics?

Kinematics is a branch of physics that studies the motion of objects without considering the forces that cause the motion.

2. What is a simple kinematics question?

A simple kinematics question involves calculating the motion of an object using basic equations of motion, such as displacement, velocity, and acceleration.

3. What are the three basic equations of motion used in kinematics?

The three basic equations of motion are:

  • Displacement (d) = Initial velocity (v0) x Time (t) + 0.5 x Acceleration (a) x Time (t)2
  • Final velocity (v) = Initial velocity (v0) + Acceleration (a) x Time (t)
  • Displacement (d) = (Initial velocity (v0) + Final velocity (v)) / 2 x Time (t)

4. How is kinematics used in real-life applications?

Kinematics is used in various fields, such as engineering, sports, and animation, to analyze and predict the motion of objects. It is used to design structures, optimize athletic performance, and create realistic movements in video games and movies.

5. What are some common misconceptions about kinematics?

One common misconception is that kinematics only applies to objects moving in a straight line. In reality, kinematics can also be applied to objects moving in a curved or circular path. Another misconception is that acceleration always means an increase in speed, when in fact it can also refer to a decrease in speed or a change in direction.

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