Look at the bottom-left parabola v = a + b*t + c*t^2. Looking at the graph, what is the value of v when t = 0? What does that say about a, b and c? What is the slope of the graph v at t = 0? What does that say about a, b and c? Finally, you have another point on the graph. That tells you more about a, b and c. You now have enough to determine the graph completely. Do something similar for the right-hand parabola, namely: use the fact that you have three points on the graph, so you can determine the three constants.AlchemistK said:Homework Statement
The graph in the attached file is a velocity-time graph.
We have to find the total displacement from t=0 to t=10.
I know that integration will be used for this problem, and i know my formulas but i do not know of how to form the equation of the two parabolas here.
I know it is of the form Ax^2 + Bx + C but how do i figure out the A.B and C?
At t = 0, the velocity is 0 and so is the slope, right? And i forgot to mention but the initial velocity is 0 so, the equation for the parabola there should be cx^2 because the vertex is at the origin.Ray Vickson said:Look at the bottom-left parabola v = a + b*t + c*t^2. Looking at the graph, what is the value of v when t = 0? What does that say about a, b and c? What is the slope of the graph v at t = 0? What does that say about a, b and c? Finally, you have another point on the graph. That tells you more about a, b and c. You now have enough to determine the graph completely. Do something similar for the right-hand parabola, namely: use the fact that you have three points on the graph, so you can determine the three constants.
RGV
Clever-Name said:Looks right to me. Now what about the 2nd parabola? Another method you can try is putting it in vertex form since you can see where the vertex is, do you remember how to do that?
The integration of the area under a v-t graph is used to calculate the displacement of an object over a specific time interval. It allows us to determine how far an object has moved, even if its velocity is constantly changing.
To integrate the area under a v-t graph, you can use the formula ∫v(t)dt = ∆x, where v(t) is the velocity function and ∆x is the displacement. You can also use the trapezoidal rule or the Simpson's rule for more accurate results.
The area under a v-t graph represents displacement, which is measured in meters (m). Therefore, the units for the area under a v-t graph are also meters (m).
Yes, the area under a v-t graph can be negative. This indicates that the object is moving in the opposite direction of the positive displacement. A negative area also means that the object is returning to its original position.
The shape of the v-t graph directly affects the area under it. A straight line (constant velocity) results in a rectangular area, while a curved line (changing velocity) results in a trapezoidal area. The steeper the slope, the greater the velocity and therefore the larger the area under the graph.