1. ## velocity graph

sorry the image is rather small. It shows the velocity of an object between time 0 and 9.

-When does the object obtain its greatest speed? I think it obtains its greatest speed at t=8, since time vs. velocity shows acceleration. Is this right

-The object was at its origin at t=3. When does it return to its origin? I thought it returned at t=6, but I got this problem wrong. Can someone guide me in the right direction because I have no clue!

2. I'm not sure what you mean by "time vs velocity shows acceleration". A "time versus velocity" graph shows velocity on its vertical axis. It is true that the slope of the graph is the accleration but that is not relevant to this problem. What is relevant is that "speed" is the absolute value of the (linear) velocity. The maximum value of the velocity shown on graph occurs at t= 5 and has value 2. The minimum value of the velocity occurs at t= 8 and is -4. But the maximum speed is |-4|= 4 so you are right about that.

The second question asks when the object is again at the origin. That means "when its position is 0", not it velocity. The information you need here is that the distance traveled is the area under the velocity graph. From t= 3 to t= 6, the velocity is positive so the object is moving in the positive direction (which I am going to call "to the right"). The graph, between t= 3 and t= 6, together with the "position= 0" axis, forms a triangle with base 6- 3= 3 and height 2. Its area is (1/2)(3)(2)= 3 so the object will move from the origin 3 places to the right. In order to get back to the origin, it must move 3 places to the left (-3).

The graph is below the axis to the right of t= 6 so the object is moving back to the right. The line passing through the points (5, 2) and (8, -4) is given by the equation
$y= \frac{-4- 2}{8- 5}= \frac{-6}{3}(t- 5)+ 2= -2(t- 5)+ 2= -2t+ 12$. If distance the object will have moved between t= 6 and t= x (which we would like to find) is the area of the right triangle having base of length x- 6, height y, and the line as hypotenuse. its area is (1/2)(x- 6)(y). Replace y with the formula above and set that area equal to -3 (it will be negative since for x> 6, y< 0- technically that's the "signed" area). Solve for x.
(Because the formula for y involves x, multiplying it by x- 5 gives a quadratic equation which may have two solutions. Of course, the one you want here is the one that is larger than 6.

3. thank you so much for that detailed response! That makes a lot of sense- thanks for covering up the gaps in my knowledge of physics