A small object travels with a velocity of v=4-t^2 where t is the time in seconds, recorded from t=0, and v is the velocity in metres/second.

a) Draw a graph of v(t) for 0¡Üt¡Ü3.

b)Find ¡Ò[0,3,(4-t^2),dt]

c) Use a suitable geomentric method to caculate the area enclosed by the graph and the x-axis for 0¡Üt¡Ü3.

d) Describe what has been calculated in (b) and (c) in terms of the journey of the object.

a) To draw the graph of v(t) for 0 ≤ t ≤ 3, we need to plot the velocity values corresponding to different time intervals.

At t = 0, the velocity is v(0) = 4 - (0)^2 = 4 m/s.
At t = 1, the velocity is v(1) = 4 - (1)^2 = 3 m/s.
At t = 2, the velocity is v(2) = 4 - (2)^2 = 0 m/s.
At t = 3, the velocity is v(3) = 4 - (3)^2 = -5 m/s.

Graphically, we'll have a curve that starts at 4 m/s, decreases, and then goes below the x-axis.

b) To find ∫[0,3,(4-t^2),dt], we need to calculate the definite integral of the function (4 - t^2) with respect to t, from t = 0 to t = 3.

∫[0,3,(4-t^2),dt] = [4t - (t^3/3)] evaluated from 0 to 3

Substituting the upper and lower limits:

= [4(3) - (3^3/3)] - [4(0) - (0^3/3)]
= [12 - 9] - [0 - 0]
= 3 - 0
= 3

Therefore, ∫[0,3,(4-t^2),dt] equals 3.

c) Using a geometric method, we need to calculate the area enclosed by the graph of v(t) and the x-axis for 0 ≤ t ≤ 3.

We recognize that the area under the graph is the sum of the areas of different shapes. In this case, we have two shapes - a rectangle and a triangle.

To calculate the area of the rectangle, we need to find the length and width. The width is the t-axis interval, which is 3 - 0 = 3. The length is the height of the rectangle, which is the maximum velocity, 4 m/s. Therefore, the area of the rectangle is length × width = 4 × 3 = 12 square units.

To calculate the area of the triangle, we need to find the base and height. The base is the same as the t-axis interval, which is 3 - 0 = 3. The height is the velocity at t = 3, which is -5 m/s. Therefore, the area of the triangle is (base × height) / 2 = (3 × -5) / 2 = -7.5 square units.

The total area enclosed by the graph and the x-axis is the sum of the areas of the rectangle and the triangle: 12 + (-7.5) = 4.5 square units.

d) In part b, we calculated the definite integral of the function (4 - t^2) with respect to t from t = 0 to t = 3. This integral value represents the net change in position or displacement of the object during the time interval from t = 0 to t = 3.

In part c, we used a geometric method to calculate the area enclosed by the graph of v(t) and the x-axis. This area represents the total distance traveled by the object during the time interval from t = 0 to t = 3, considering both positive and negative displacements.

So, in terms of the journey of the object, part b calculated the net change in position, and part c calculated the total distance traveled.