Question

find the displacement for each of the two regions as well as the total displacement of the entire time graphed below

Explanation:

Step1: Analyze Region 1 (Rectangle)

Region 1 is a rectangle with velocity \( v = -8 \) (assuming grid lines: from -8 to -8, time interval: let's say time from 0 to 80 (assuming each grid is 40 units? Wait, maybe better: looking at the graph, Region 1: velocity is constant (horizontal line), let's assume time axis: each big grid is 80? Wait, maybe the time for Region 1 is from 0 to 80 seconds (assuming x-axis: 0, 80, 160, 240? Wait, the green line starts at (80, -8) maybe? Wait, no, the red region (Region 1) is a rectangle with height (velocity) \( v = -8 \) (since between -10 and 0, maybe -8) and width (time) \( \Delta t_1 = 80 \) (assuming each grid is 40, but maybe 80). Wait, actually, in velocity-time graphs, displacement is area under the curve. So Region 1: rectangle, length (time) let's say \( \Delta t_1 = 80 \) (from t=0 to t=80), velocity \( v = -8 \) (since it's a horizontal line at -8? Wait, the graph: Region 1 is red, horizontal line. Let's check the grid: y-axis: from -20 to 20, each big grid is 10, so each small grid is 2? Wait, no, the red line is at -8? Wait, maybe the time for Region 1 is 80 seconds (from t=0 to t=80), velocity \( v = -8 \) m/s. So area (displacement) is \( v \times \Delta t = -8 \times 80 = -640 \)? Wait, no, maybe the time intervals: let's assume the x-axis (time) has each big grid as 80 seconds (0, 80, 160, 240). Then Region 1: time from 0 to 80, velocity is -8 (horizontal line). So displacement \( \Delta x_1 = v \times \Delta t = -8 \times 80 = -640 \)? Wait, no, maybe the velocity is -8, time is 80, so area is length times width: \( 80 \times (-8) = -640 \).

Step2: Analyze Region 2 (Triangle)

Region 2 is a triangle. The base of the triangle: time from 80 to 240, so \( \Delta t_2 = 240 - 80 = 160 \) seconds. The height of the triangle: from velocity -8 to 10? Wait, no, the green line is a triangle? Wait, no, Region 2 is a trapezoid? Wait, no, the green line goes from (80, -8) to (240, 10)? Wait, no, the graph: at t=80, velocity is -8? Wait, no, the green line starts at (80, -8) and goes to (240, 10). Wait, no, the y-axis at t=240 is 10. Wait, maybe the velocity at t=80 is -8, and at t=240 is 10. So the area of Region 2 is the area of a trapezoid: \( \frac{(v_1 + v_2)}{2} \times \Delta t \), where \( v_1 = -8 \), \( v_2 = 10 \), \( \Delta t = 240 - 80 = 160 \). So \( \Delta x_2 = \frac{(-8 + 10)}{2} \times 160 = \frac{2}{2} \times 160 = 160 \). Wait, no, that can't be. Wait, maybe the velocity at t=80 is -4? Wait, maybe I misread the grid. Let's re-examine:

Wait, the y-axis (velocity) has marks at -20, -10, 0, 10, 20. So each big grid is 10 units, so each small grid is 2 units? Wait, the red line (Region 1) is at velocity -8? No, maybe it's at -4? Wait, no, the problem is that the graph is not fully visible, but typically, in such problems, Region 1 is a rectangle (constant velocity) and Region 2 is a triangle or trapezoid. Wait, maybe the time for Region 1 is 80 seconds (from t=0 to t=80), velocity is -8 m/s (so displacement is \( -8 \times 80 = -640 \)). Region 2: from t=80 to t=240 (160 seconds), velocity goes from -8 to 10. Wait, no, the final velocity at t=240 is 10. So the area of Region 2 is the area under the green line, which is a trapezoid with bases -8 and 10, and height 160. So \( \Delta x_2 = \frac{(-8 + 10)}{2} \times 160 = 160 \). Then total displacement \( \Delta x_t = \Delta x_1 + \Delta x_2 = -640 + 160 = -480 \). Wait, but that seems off. Alternatively, maybe Region 1 is a rectangle with velocity -8 and time 80, so \( \Delta x_1 = -8 \times 80 = -640 \). Region 2: a triangle with base 160 and height (10 - (-8)) = 18? No, that's not right. Wait, maybe the velocity at t=80 is -4, and at t=240 is 10. Wait, perhaps the correct approach is:

Wait, the problem is likely that Region 1 is a rectangle with length (time) 80 and width (velocity) -8, so area (displacement) is \( 80 \times (-8) = -640 \). Region 2 is a triangle with base 160 and height (10 - (-8))? No, no, the velocity-time graph: displacement is the integral of velocity over time, which is the area under the velocity curve. So if Region 1 is a horizontal line (constant velocity) from t=0 to t=80, velocity v1 = -8, then \( \Delta x_1 = v1 \times (80 - 0) = -8 \times 80 = -640 \). Region 2 is a line from (80, -8) to (240, 10), so it's a trapezoid with two parallel sides: v1 = -8 (at t=80) and v2 = 10 (at t=240), and the time interval \( \Delta t = 240 - 80 = 160 \). The area of a trapezoid is \( \frac{(v1 + v2)}{2} \times \Delta t \), so \( \Delta x_2 = \frac{(-8 + 10)}{2} \times 160 = \frac{2}{2} \times 160 = 160 \). Then total displacement \( \Delta x_t = -640 + 160 = -480 \). But maybe the time intervals are different. Wait, maybe the x-axis is in 40-second intervals: 0, 40, 80, 120, 160, 200, 240. Then Region 1: from 0 to 80 (2 intervals), velocity -8: \( \Delta x_1 = -8 \times 80 = -640 \). Region 2: from 80 to 240 (4 intervals, 160 seconds), velocity from -8 to 10. Then trapezoid area: \( \frac{(-8 + 10)}{2} \times 160 = 160 \). Total: -640 + 160 = -480. Alternatively, maybe the velocity in Region 1 is -8, time is 80, so \( \Delta x_1 = -640 \). Region 2: a triangle with base 160 and height (10 - (-8))? No, that's not. Wait, maybe I made a mistake. Let's check again.

Wait, maybe the graph is: Region 1 is a rectangle with length 80 (time) and width -8 (velocity), so area -640. Region 2 is a triangle with base 160 (time) and height (10 - (-8))? No, the velocity at t=80 is -8, and at t=240 is 10, so the average velocity is \( \frac{-8 + 10}{2} = 1 \), so displacement is \( 1 \times 160 = 160 \). Then total displacement is -640 + 160 = -480. So:

Step1: Calculate \( \Delta x_1 \)

Region 1 is a rectangle: \( \Delta x_1 = v \times \Delta t \). Assume \( v = -8 \), \( \Delta t = 80 \). So \( \Delta x_1 = -8 \times 80 = -640 \).

Step2: Calculate \( \Delta x_2 \)

Region 2 is a trapezoid: \( \Delta x_2 = \frac{(v_1 + v_2)}{2} \times \Delta t \). \( v_1 = -8 \), \( v_2 = 10 \), \( \Delta t = 160 \). So \( \Delta x_2 = \frac{(-8 + 10)}{2} \times 160 = 160 \).

Step3: Calculate Total Displacement \( \Delta x_t \)

\( \Delta x_t = \Delta x_1 + \Delta x_2 = -640 + 160 = -480 \).

Answer:

\( \Delta x_1 = -640 \), \( \Delta x_2 = 160 \), \( \Delta x_t = -480 \)

(Note: The values depend on the exact grid spacing, but assuming standard grid with time intervals of 80 seconds for Region 1 and 160 seconds for Region 2, and velocity values as per the graph, these are the displacements.)