during which time interval does the object travel approximately 10 meters?\na. 0 seconds to 3 seconds\nb. 3…

Explanation:

Step1: Recall the formula for distance

The distance traveled by an object in a velocity - time graph is given by the area under the curve. For a linear segment of the graph, the area of the trapezoid (if non - horizontal) or rectangle (if horizontal) formed by the velocity - time curve, the time - axis, and the vertical lines at the start and end of the time interval is calculated using the appropriate area formula.

Step2: Calculate area for option A

For the time interval from 0 to 3 seconds: The velocity at (t = 0) is (v_1=10\ m/s) and at (t = 3) is (v_2 = 6\ m/s). The area of the trapezoid (A=\frac{(v_1 + v_2)}{2}\times\Delta t), where (\Delta t=3\ s), (v_1 = 10\ m/s), (v_2=6\ m/s). So (A=\frac{(10 + 6)}{2}\times3=\frac{16}{2}\times3=24\ m).

Step3: Calculate area for option B

For the time interval from 3 to 5 seconds: The velocity at (t = 3) is (v_1 = 6\ m/s) and at (t = 5) is (v_2=5\ m/s). (\Delta t = 2\ s). The area of the trapezoid (A=\frac{(v_1 + v_2)}{2}\times\Delta t=\frac{(6 + 5)}{2}\times2=11\ m\approx10\ m).

Step4: Calculate area for option C

For the time interval from 5 to 7 seconds: The velocity is constant at (v = 5\ m/s), (\Delta t=2\ s). The area (rectangle) (A=v\times\Delta t=5\times2 = 10\ m).

Step5: Calculate area for option D

For the time interval from 7 to 8 seconds: The velocity at (t = 7) is (v_1 = 5\ m/s) and at (t = 8) is (v_2=3\ m/s), (\Delta t = 1\ s). The area of the trapezoid (A=\frac{(v_1 + v_2)}{2}\times\Delta t=\frac{(5+3)}{2}\times1 = 4\ m).

Step6: Calculate area for option E

For the time interval from 8 to 10 seconds: The velocity at (t = 8) is (v_1 = 3\ m/s) and at (t = 10) is (v_2=0\ m/s), (\Delta t = 2\ s). The area of the trapezoid (A=\frac{(v_1 + v_2)}{2}\times\Delta t=\frac{(3 + 0)}{2}\times2=3\ m).

Answer:

C. 5 seconds to 7 seconds