Solving Displacement of a Ball in 5.0-15.0 Time Interval

[Throwback24]

Homework Statement

A balls roll along a straight path with the following velocity and time intervals

3.0 s at +15.0 m/s
7.0 s at +20 m/s
6.0 s at -30 m/s
4.0 s at 3.5 m/s

What is the displacement in the last 5.0-15.0 time interval:

I've been looking at this question for a good 30 mins, I don't know where to start :(

 

[mgb_phys]

1, graph paper.
2, Mark seconds along the bottom,distance on side
3, Each part is a slope the 15 m/s means you go up 15m for every 1sec you go along
4, The total distance at any time can be read from the graph

Once you have done this it's pretty easy to see how you can do it arithmeticaly

 

[LowlyPion]

Consider if you drew a graph of this data. Along the x - axis if you have time and along the y-axis you have velocity. You will have a graph with 4 velocity levels looking like a histogram, through the range of time given to 20 seconds.

Now if you look at the area under this curve, each square of area (1 second along the x, and 1 m/s along the y) yields what? Won't it be 1m/s*1s = 1m?

So won't it be that if you count up each of those squares under the curve between 5 s and 15s that that total will be the total displacement?

Welcome to integration.

 

[PhanthomJay]

Homework Statement

A balls roll along a straight path with the following velocity and time intervals

3.0 s at +15.0 m/s
7.0 s at +20 m/s
6.0 s at -30 m/s
4.0 s at 3.5 m/s

What is the displacement in the last 5.0-15.0 time interval:

I've been looking at this question for a good 30 mins, I don't know where to start :(

For the constant velocity during each time interval, use displacement = (velocity)*(time). Where is the ball at 3 seconds, traveling at v=15? at 5 seconds, traveling at v = 20? at 10 seconds, traveling ay v =20? at 15 seconds, traveling at v = -30? Now calculate where it was at 5 and where it was at 15, and note the difference in the dispalacement, and the direction of the displacement.

 

[Throwback24]

Oh! I get it. Thank you very much.

I'm trying to graph a Position-Time graph, Velocity-Time graph, and Acceleration-Time graph but am stuck.

I get what the question is asking me but aren't all graphs going to look exactly the same minus what I name the x and y axis?

Here's the Q:

You are riding in an elevator. Starting from rest, the eleveator undergoes the following motions.

It accelerates from rest (v=0 m/s) upwards for 5 s at 2 m/s2
It then coasts for 20 s at 10 m/s (a=0 m/s2)
Finally, starting at 10 /s, it accelerates downward for 2.5 s at -4 m/s2.

Wouldn't they all be the same? I drew my points for the position time graph. But my other two graphs look the same :S

 

[LowlyPion]

No. All your graphs will not be the same.

Acceleration is going to represent the slope of the velocity graph at all points of time and the velocity graph is going to represent the slope of the position graph at all points of time. Acceleration if you will recall is the rate of change of velocity - the tangent at each point. And Velocity is the rate of change of position. They are derivatives.

Alternatively you can start with the statement of the problem and draw your acceleration graph that will show the value of acceleration - a constant for the first 5 seconds - a horizontal line on the graph. On the Velocity graph each point represents the definite integral from 0 to that point in time and it will appear on the velocity graph as a sloped line, whose slope would be the value during that period from the acceleration graph. Likewise again for the position graph being the integral of velocity, except that this time the graph will be a quadratic curve (think parabola) proportional to position squared (x² ).