I'm not sure what you mean by "time vs velocity shows acceleration". A "time versus velocity" graph showsvelocityon its vertical axis. It is true that theslopeof 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 thevelocityshown 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 maximumspeedis |-4|= 4 so you are right about that.

The second question asks when the object is again at theorigin. That means "when itspositionis 0", not it velocity. The information you need here is that the distance traveled is theareaunder 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

. 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.