Question

1. At what times does the object change direction? How do you know?
2. What is the object doing at t = 6 s and t = 9 s. (Hint: It’s not changing direction.)
3. In which direction is the object moving at t=7s? Explain.
4. At what time is the object farthest from its initial position? How do you know?

Answer 1

1) There is a corner.
2) Look at it. Up, down, dancing the jig?
3) Isn't that between 6 and 9?
4) Looks like right around 4 it starts heading back where it came from.

Answer 2

what do you mean there is a corner, is it at t=2?

Answer 3

Object changes direction when its velocity went from positive to negative, or negative to positive. Thinking of car slowing down until it goes backward or the other way around.

Answer 4

so from t=4 to t=8

Answer 5

From t = 0, to t = 2, the object is moving with constant velocity. Then the object begins to slow down. The velocity becomes zero when t = 4. Then for t > 4, the velocity is negative. That means the object is moving in the opposite direction. Then it changes direction again at t = 8.

Answer 6

1. At what times does the object change direction? How do you know?
At t = 4 seconds and at t = 8 seconds.
A little before t = 4 seconds, the object has a positive velocity and therefore it is moving along the positive x-direction. A little after t = 4 seconds, the object has a negative velocity which indicates the object is moving along the negative x direction. The object has changed direction.
At t = 8 seconds it changes direction from negative x to positive x.

Answer 7

Answer to 1 is not clear?

Answer 8

so for 2. is it that it is increasing?

Answer 9

A little to the left of t = 6, the slope of the velocity curve is negative which implies the object is decelerating. A little after t = 6, the slope of the velocity curve is positive which implies the object is accelerating.
At t = 9 s, it changes from acceleration to deceleration.

Answer 10

would 3. be is moving in the negative direction. and 4. at t=9

Answer 11

Yes, #3: Object is moving in the negative direction.

Answer 12

For #4, From the definition of velocity, v = dx/dt
Therefore, distance traveled, x = integral of the velocity which can be computed graphically by summing up the area under the curve. As t goes from 0 to 2 to 4, the area under the curve keeps on increasing. So at t = 4 seconds, the object is the farthest. Then, as t increases to 6 and then to 8, the area under the curve becomes negative and starts to decrease the total area. Then the area under the curve starts to increase again. But the maximum area under the curve is attained when t = 4 seconds.

Answer 13

If you have not been taught integration and area under the curve yet you can answer this question as follows: The object has positive velocity until t = 4 seconds. After that the velocity becomes negative implying the object has turned around and moving in the direction of where it started. It keeps on doing that until t = 8. Then it starts to move away but not for very long. The max distance away happens when t = 4.

Answer 14

My bad. The suggestion for #1

Anywhere the curve is not FLAT is a change in speed.
Anywhere the curve crosses the x-axis is a change in DIRECTION