a particle moves along the x-axis so that its position at time 0 ≤ t ≤ 18 is given by x(t), whose graph is…

Explanation:

Step1: Analyze intervals of motion

The total distance traveled is the sum of the absolute values of the displacements over each interval where the particle changes direction (i.e., where the velocity changes sign, corresponding to the position graph having local maxima or minima, or crossing the t - axis). We identify the key points on the graph: at (t = 0), (x(0)=6); local minimum around (t = 5), local maximum around (t = 8), crosses the t - axis at (t = 15), local minimum around (t = 17), and at (t = 18), (x(18)=3). We break the interval ([0,18]) into sub - intervals: ([0,5]), ([5,8]), ([8,15]), ([15,17]), ([17,18]).

Step2: Calculate displacement on ([0,5])

At (t = 0), (x(0)=6); at (t = 5), from the graph, (x(5)=2). The displacement is (x(5)-x(0)=2 - 6=- 4). The distance is (| - 4|=4).

Step3: Calculate displacement on ([5,8])

At (t = 5), (x(5)=2); at (t = 8), from the graph, (x(8)=6). The displacement is (x(8)-x(5)=6 - 2 = 4). The distance is (|4| = 4).

Step4: Calculate displacement on ([8,15])

At (t = 8), (x(8)=6); at (t = 15), (x(15)=0) (since it crosses the t - axis). The displacement is (x(15)-x(8)=0 - 6=-6). The distance is (|-6| = 6).

Step5: Calculate displacement on ([15,17])

At (t = 15), (x(15)=0); at (t = 17), from the graph, (x(17)=- 2). The displacement is (x(17)-x(15)=-2-0=-2). The distance is (|-2| = 2).

Step6: Calculate displacement on ([17,18])

At (t = 17), (x(17)=-2); at (t = 18), (x(18)=3). The displacement is (x(18)-x(17)=3-(-2)=5). The distance is (|5| = 5).

Step7: Sum the distances

Now, we sum up all the distances: (4 + 4+6 + 2+5=21)? Wait, no, let's re - check the graph values more accurately. Let's use the grid. Each grid square is 1 unit.

• From (t = 0) to (t = 5): The particle moves from (x = 6) to (x = 2). The vertical change is (6 - 2=4), distance (= 4).
• From (t = 5) to (t = 8): Moves from (x = 2) to (x = 6). Vertical change (=6 - 2 = 4), distance (= 4).
• From (t = 8) to (t = 15): Moves from (x = 6) to (x = 0). Vertical change (=6-0 = 6), distance (= 6).
• From (t = 15) to (t = 17): Moves from (x = 0) to (x=-2). Vertical change (=0-\left(-2\right)=2)? Wait, no, from (x = 0) to (x=-2), the displacement is (- 2), distance (= 2).
• From (t = 17) to (t = 18): Moves from (x=-2) to (x = 3). Vertical change (=3-\left(-2\right)=5), distance (= 5). Wait, maybe my initial key points were wrong. Let's re - evaluate with more precise grid reading.

Wait, another approach: The total distance is the integral of (|v(t)|) from (0) to (18), and since (v(t)=x^\prime(t)), the total distance is the sum of the lengths of the vertical segments between the peaks and troughs.

Looking at the graph:

• From (t = 0) (x = 6) to the first minimum (t≈5, x = 2): distance is (6 - 2=4).
• From t≈5 (x = 2) to the first maximum (t≈8, x = 6): distance is (6 - 2 = 4).
• From t≈8 (x = 6) to the t - axis (t = 15, x = 0): distance is (6-0 = 6).
• From t = 15 (x = 0) to the second minimum (t≈17, x=-2): distance is (0-\left(-2\right)=2)? No, distance is (|0-\left(-2\right)|=2)? Wait, no, when moving from (x = 0) to (x=-2), the distance is (2) (since it goes down 2 units).
• From t≈17 (x=-2) to t = 18 (x = 3): distance is (3-\left(-2\right)=5).

Wait, but let's check the final position. At (t = 18), (x = 3). Let's also consider the correct way:

Alternative way:

1. From (t = 0) to the first minimum (let's say (t = 5)): starts at 6, goes to 2. Distance: (6 - 2=4).
2. From (t = 5) to the first maximum ( (t = 8)): goes from 2 to 6. Distance: (6 - 2 = 4).
3. From (t = 8) to crossing the t - axis ( (t = 15)): goes from 6 to 0. Distance: (6-0 = 6).
4. From (t = 15) to the second minimum ( (t = 17)): goes from 0 to - 2. Distance: (0-\left(-2\right)=2) (absolute value, so 2).
5. From (t = 17) to (t = 18): goes from - 2 to 3. Distance: (3-\left(-2\right)=5).

Now sum them up: (4 + 4+6 + 2+5=21)? Wait, no, maybe I made a mistake in the second minimum. Wait, when (t = 17), the y - value is - 2, and at (t = 18), it's 3. But let's check the total displacement and then adjust. Wait, the correct way is to look at the graph's vertical changes:

Wait, let's count the units:

• From (0,6) to (5,2): vertical difference is 4 (down 4).
• From (5,2) to (8,6): vertical difference is 4 (up 4).
• From (8,6) to (15,0): vertical difference is 6 (down 6).
• From (15,0) to (17, - 2): vertical difference is 2 (down 2).
• From (17, - 2) to (18,3): vertical difference is 5 (up 5).

Now sum: (4 + 4+6 + 2+5=21)? Wait, but let's check with another method. The total distance is also equal to the sum of the lengths of the "ups" and "downs".

Wait, maybe the correct answer is 21? Wait, no, let's re - examine the graph. Let's assume the grid is 1 unit per square.

At (t = 0), (x = 6).

At (t = 5), (x = 2): distance from 0 to 5: (|6 - 2|=4).

At (t = 8), (x = 6): distance from 5 to 8: (|2 - 6|=4).

At (t = 15), (x = 0): distance from 8 to 15: (|6 - 0|=6).

At (t = 17), (x=-2): distance from 15 to 17: (|0-\left(-2\right)|=2).

At (t = 18), (x = 3): distance from 17 to 18: (|-2 - 3|=5) (wait, no, distance is (|3-\left(-2\right)| = 5)).

Now sum: (4 + 4+6 + 2+5=21). Wait, but maybe I missed a segment. Wait, another way: the total distance is the sum of all the "arcs" length in terms of vertical change.

Wait, maybe the correct answer is 21. But let's check again.

Wait, when the particle moves from (x = 6) (t = 0) to (x = 2) (t = 5): distance 4.

From (x = 2) (t = 5) to (x = 6) (t = 8): distance 4.

From (x = 6) (t = 8) to (x = 0) (t = 15): distance 6.

From (x = 0) (t = 15) to (x=-2) (t = 17): distance 2.

From (x=-2) (t = 17) to (x = 3) (t = 18): distance 5.

(4 + 4+6+2 + 5=21).

Answer:

(21)