The times in the past when continental glaciers covered large parts of Earth’s land surface were the ____________________.
2 answers:
You might be interested in
Answer:
a. The bug is moving at a constant speed of -2 m/s
b. The total distance traveled by the bug is 17 meters
c. The distance the bug end up from its starting point is 3 meters
d. i. The bugs velocity is 0 when t = 0, 5, or 10 seconds,
ii. The bugs velocity is > 0 at 5 ≤ t ≤ 10
iii. The bugs velocity is < 0 at 0 ≤ t ≤ 5
e. i. The bug's acceleration is 0 at 2 ≤ t ≤ 4
ii. The bug's acceleration is > 0 at 4 ≤ t ≤ 6 and 8 ≤ t ≤ 9
iii. The bug's acceleration is > 0 at 0 ≤ t ≤ 2 and 9 ≤ t ≤ 10
f. The time the bug passes the starting point is approximately 8.83 seconds
Explanation:
a. When 2 ≤ t ≤ 4, the bug is moving at a constant speed of -2 m/s
b. The total distance the bug traveled is given by the area under the velocity-time graph which consist of two trapezoid and a triangle and is therefore is given as follows;
The area of the first (negative) trapezoid = -2 × (2 + 5)/2 = -7
The area of the second trapezoid = 2 × (3.5 + 5)/2 = 8.5
The area of the triangle = 1/2 × 1.5 × 2 = 1.5
The sum of the magnitude of the areas = The total distance traveled by the bug = + 8.5 + 1.5 = 17
The total distance traveled by the bug = 17 meters
c. The distance the bug end up from its starting point is given by the sum of the displacement (total displacement) of the bug
The total displacement of the bug = -7 + 8.5 + 1.5 = 3
The distance the bug end up from its starting point = The total displacement of the bug = 3 meters
d. i. The bugs velocity is 0 when the graph crosses the time, 't', axis, which is at points for time, t = 0 seconds, t = 5 seconds, and t = 10 seconds
ii. The bugs velocity is > 0 between the times, 5 ≤ t ≤ 10
iii. The bugs velocity is < 0 between the times, 0 ≤ t ≤ 5
e. i. The bug's acceleration is 0 between the times, 2 ≤ t ≤ 4
ii. The bug's acceleration is > 0 between the times, 4 ≤ t ≤ 6 and 8 ≤ t ≤ 9
iii. The bug's acceleration is > 0 between the times, 0 ≤ t ≤ 2 and 9 ≤ t ≤ 10
f. The bug passes the starting point at the time, 8 + t, where 't', is given by the area of a trapezoid as follows;
2 = t × (2 + 2 + t)/2
∴ 4 = 4·t + t²
t² + 4·t - 4 = 0
(t + (2 + 2·√2))·(t - (2·√2 - 2))
∴ t = 2·√2 - 2
The time the bug passes the starting point = 8 + 2·√2 - 2 = 6 + 2·√2 ≈ 8.83
The time the bug passes the starting point ≈ 8.83 seconds
