Question:

An object is subjected to one-dimensional motion along the x-axis. The position-time graph of the object is shown below:

1. Determine the velocity of the object for each of the following time intervals:

   a) Interval AB b) Interval BC c) Interval CD

2. Determine the acceleration of the object for each of the following time intervals:

   a) Interval AB b) Interval BC c) Interval CD

3. Determine the total distance covered by the object during the entire motion.

Answer:

1. Determining velocity:

a) Interval AB:

To determine the velocity during interval AB, we need to find the slope of the position-time graph. The slope of a distance-time graph represents the rate of change of distance, which is equivalent to velocity.

From the graph, we can see that during interval AB, the position remains constant, indicating that the object is at rest. Therefore, the velocity during interval AB is zero.

b) Interval BC:

During interval BC, the position-time graph is a straight line with a positive slope. The slope of this line gives us the velocity of the object during this interval.

To calculate the slope, we can select two points on the line. Let's choose point B (t=0 s, x=-15 m) and point C (t=3 s, x=10 m).

The velocity (v) can be calculated using the formula:

v = Δx / Δt,

where Δx is the change in position and Δt is the change in time.

Δx = x2 - x1 = 10 m - (-15 m) = 25 m, Δt = t2 - t1 = 3 s - 0 s = 3 s.

Now, substituting the values in the formula:

v = Δx / Δt = 25 m / 3 s ≈ 8.33 m/s.

Therefore, the velocity during interval BC is approximately 8.33 m/s.

c) Interval CD:

During interval CD, the position-time graph is a horizontal line, indicating that the object is at rest. Hence, the velocity during interval CD is zero.

2. Determining acceleration:

a) Interval AB:

The acceleration can be determined using the slope of the velocity-time graph. As the velocity remains constant (zero) during interval AB, it means the acceleration is also zero.

b) Interval BC:

To determine the acceleration during interval BC, we need to find the slope of the velocity-time graph. We can use the same points B and C mentioned earlier and the formula:

a = Δv / Δt,

where Δv is the change in velocity and Δt is the change in time.

Δv = v2 - v1 = 8.33 m/s - 0 m/s = 8.33 m/s, Δt = t2 - t1 = 3 s - 0 s = 3 s.

Substituting the values:

a = Δv / Δt = 8.33 m/s / 3 s ≈ 2.78 m/s².

Therefore, the acceleration during interval BC is approximately 2.78 m/s².

c) Interval CD:

Since the velocity remains constant (zero) during interval CD, the acceleration is also zero.

3. Total distance covered:

To determine the total distance covered by the object, we need to calculate the area under the position-time graph.

For interval AB, the area is a rectangle with a height of 0 m and a base of 4 s, so the area is 0 m.

For interval BC, the area is a trapezoid. The formula for the area of a trapezoid is given by:

A = (b1 + b2) * h / 2,

where b1 and b2 are the parallel sides, and h is the height.

Using the points B (-15 m, 0 s) and C (10 m, 3 s):

b1 = -15 m, b2 = 10 m, h = 3 s.

Substituting the values:

A = (b1 + b2) * h / 2 = (-15 m + 10 m) * 3 s / 2 = -22.5 m·s.

For interval CD, the area is a rectangle with a height of 10 m and a base of 4 s, so the area is 40 m.

Therefore, the total distance covered by the object is given by the sum of these areas:

Total distance = 0 m + (-22.5 m·s) + 40 m = 17.5 m.

Thus, the object covers a total distance of 17.5 meters during the entire motion.