An object starts from rest at the origin and moves along the x axis with a constant acceleration of 4m/s^2. Its average velocity as it goes from x=2m to x=8m is:

A. 1m/s
B. 2m/s
C. 3m/s
D. 5m/s
E. 6m/s

An object is released from rest and falls a distance H during the first second of time. How far will it fall during the next secondof time?

A. H
B. 2H
C. 3H
D. 4H
E. H^2

If an abject has a constatn acceleration of 3m/s^2. The displacement versus time graph for this object has a slope:?

A. that increases with time
B. that is constant
C. that decreases with time
D. of 3 m/s
E. of 3 m/s^2

What is your thinking on these? I am not much inclined to do them for you, but will be happy to critique thinking.

1.e

1a 2b 3c

To find the average velocity of an object as it goes from one position to another, we can use the formula:

average velocity = (final position - initial position) / time

For the first question, the object starts from rest at the origin, so its initial position is 0m and its final position is 8m. The time is not given, but since the object has a constant acceleration, we can use the formula:

final velocity = initial velocity + (acceleration * time)

Since the object starts from rest, the initial velocity is 0m/s. We can rearrange this formula to solve for time:

time = (final velocity - initial velocity) / acceleration

We know the acceleration is 4m/s^2, and the final velocity is the average velocity. So we can substitute the values into the equation:

time = (average velocity - 0) / 4

Simplifying, we get:

time = average velocity / 4

Now we can substitute this value of time into the first formula to find the average velocity:

average velocity = (8 - 0) / (average velocity / 4)

Simplifying further:

average velocity = (8 * 4) / average velocity

average velocity = 32 / average velocity

Multiplying both sides by average velocity:

average velocity^2 = 32

Taking the square root of both sides:

average velocity = √32

Since 32 can be simplified as 4 * 8, we have:

average velocity = √(4 * 8)

average velocity = √4 * √8

average velocity = 2 * √8

So the average velocity as the object goes from x=2m to x=8m is approximately 2 * 2.83 = 5.66m/s.

Therefore, the correct answer is D. 5m/s.

For the second question, when an object is released from rest and falls a distance H during the first second of time, we can use the equations of motion to find the distance it will fall during the next second.

The equation for the position of a falling object is:

position = initial position + (initial velocity * time) + (0.5 * acceleration * time^2)

Since the object is released from rest, the initial velocity is 0m/s. The acceleration due to gravity is approximately 9.8m/s^2. In the first second, time is 1 second, and the distance is H. So we can substitute these values into the equation:

H = 0 + (0 * 1) + (0.5 * 9.8 * 1^2)

H = 0 + 0 + (0.5 * 9.8 * 1)

H = 0 + 0 + 4.9

H = 4.9

Therefore, the object will fall a distance of 4.9 meters during the next second of time.

Thus, the correct answer is D. 4H.

For the third question, the slope of a displacement versus time graph represents the velocity of an object. Since the acceleration is constant at 3m/s^2, the velocity increases linearly with time.

Therefore, the correct answer is A. that increases with time.

I hope this helps! Let me know if you have any further questions.