I'm kind of rusty on my calc here...
ok ther's a graph of velocity vs time w/ velocity in m/s and time in seconds were time is of course the x-axis
k I'm asked to describe how the velocity changes over the timer interval in this interval
the graph consists of three segments joined together
i said that for the first segment which is slanted upward and is a straight line
dy/dx = 5 m/s for the time interval of the segment
i said for the second segment which is completely flat straight line that
dy/dx = 0 m/s for the time interval of the segment
I said for the third segment a completley straight line slanted downwards that
dy/dx = -2.5 m/s
ok i think that's the right notation correct?
Then I was asked how does the distance traveled by the car of the interval of the segment relate to the velocity of the graph shown
ok so im asked to find the area of a rectangular area I forget the notation for this
X - Xo = INTEGRAL
were the top number of the integral is the last x value of the second line segment
the bottom number of the integral is the first x value of the second line segment right?
now what goes to the right of the integral i forget this...
were do i put the y value of the line...
basically how do i right this in calculus notation and then simplify it
a line segment, were dy/dx = 0 at y = 20 from the interval x = 16 and x = 4
i know that it simplifies to
(16-4)20 but they wont the calculus notation first and then show the simplified version...
thannks...
To describe the velocity changes over the time interval in the given graph, you correctly determined the values of dy/dx (which represents the rate of change of velocity with respect to time) for each segment.
For the first segment, where the graph is a straight line slanted upward, you correctly calculated dy/dx = 5 m/s.
For the second segment, which is a completely flat straight line, the velocity remains constant, so dy/dx = 0 m/s.
For the third segment, which is a straight line slanted downwards, you correctly determined dy/dx = -2.5 m/s.
Now, moving on to the distance traveled by the car during each segment and its relationship to the velocity graph:
To find the distance traveled by the car during an interval, we need to calculate the area under the velocity graph within that interval.
In calculus notation, the area under a curve is found using integration. Specifically, for the interval between x = a and x = b, the notation is:
∫(from a to b) y dx
In this case, since the velocity graph provided has a constant velocity during the second segment, the velocity is 0 m/s. Therefore, the distance traveled during this segment is simply the velocity multiplied by the time interval, which is 0 m/s multiplied by the interval length, giving us zero distance.
For the first and third segments, which are straight lines, we can determine the distance traveled by finding the area of the corresponding rectangles.
For the first segment, where dy/dx = 5 m/s, the height of the rectangle is 5 m/s and the time interval is given by the difference between the x-values at the start and end of the segment. Therefore, the distance traveled during this segment is:
∫(from x1 to x2) 5 dx
Integrating with respect to x, we get:
(5x) | (from x1 to x2) = (5 * x2) - (5 * x1)
Similarly, for the third segment, where dy/dx = -2.5 m/s, the height of the rectangle is -2.5 m/s. The distance traveled during this segment is:
∫(from x3 to x4) (-2.5) dx
Integrating, we get:
(-2.5x) | (from x3 to x4) = (-2.5 * x4) - (-2.5 * x3)
Simplifying these expressions will give you the final results for the distances traveled during each segment.