Question

question 1 1 pts the graph represents the motion of an object over an interval of time in a straight line. starting from rest at time t = 0. the object spends three seconds moving at its maximum speed of 189 meters per second. find the acceleration (in m/s²) of the object at the moment when it comes to rest during this interval. provide at least one decimal place

Explanation:

Step1: Calculate total distance

First, we find the total distance the object travels, which is the area under the v-t graph. The graph is a trapezoid (or we can split it into a rectangle for the constant speed part, and two triangles for acceleration/deceleration, plus the final deceleration triangle). But we know:
1. The object starts from rest, accelerates to max speed $v_{max}=189$ m/s, travels at this speed for 3 seconds, then decelerates to rest.
Let the time to accelerate be $t_1$, time to decelerate to rest be $t_3$, and constant speed time $t_2=3$ s.
Total distance $d = \frac{1}{2}v_{max}t_1 + v_{max}t_2 + \frac{1}{2}v_{max}t_3$
But we also know that the area under the v-t graph is equal to the total displacement, and for the entire motion, the change in velocity from start to end is 0, so the area of the acceleration triangle equals the area of the deceleration triangles? No, wait, no—wait, the final part is decelerating to rest, so the last segment is the deceleration to rest. Wait, no, let's re-express:
Wait, the problem says we need the acceleration when it comes to rest, which is the slope of the final segment (deceleration to rest). Let's denote:
Let the time taken to accelerate to 189 m/s be $t_1$, time at constant speed is 3 s, time to decelerate to the lower constant speed? No, wait no—the graph: starts at 0, accelerates to a speed, stays constant, decelerates to another constant? No, no, the problem says "starting from rest at t=0, reaches max speed 189 m/s, spends 3 seconds at that speed, then comes to rest". So the v-t graph is:
1. Line from (0,0) to $(t_1, 189)$: acceleration phase
2. Horizontal line from $(t_1,189)$ to $(t_1+3, 189)$: constant speed
3. Line from $(t_1+3,189)$ to $(t_1+3+t_3, 0)$: deceleration to rest

But wait, the total distance is the area under the graph:
$d = \frac{1}{2} \times 189 \times t_1 + 189 \times 3 + \frac{1}{2} \times 189 \times t_3$
But we also know that the change in velocity for acceleration is $189 = a_1 t_1$, and for deceleration, $0 = 189 + a_3 t_3$ (so $a_3 = -\frac{189}{t_3}$). But wait, no—wait, the problem says "the graph represents the motion... starting from rest at t=0. The object spends three seconds moving at its maximum speed of 189 meters per second. Find the acceleration (in $\frac{m}{s^2}$) of the object at the moment when it comes to rest during this interval."

Wait, no—wait, the v-t graph: the area under the graph is the total distance, but we are missing the total time? No, wait, no—wait, the graph is a trapezoid? No, wait, looking at the graph: it has 4 segments:
1. Acceleration from 0 to $v_1$
2. Constant speed $v_1$
3. Deceleration to $v_2$
4. Constant speed $v_2$
5. Deceleration to 0? No, no, the problem says maximum speed is 189, so the highest vertical line is 189. So the top horizontal segment is 189 m/s, duration 3 s.

Wait, let's count the grid squares:
- The constant speed segment (top horizontal) is 3 squares wide, which equals 3 seconds. So each grid square on the t-axis is 1 second.
- The maximum speed is 189 m/s, which is 7 grid squares tall (count the vertical squares for the top segment). So each vertical square is $\frac{189}{7}=27$ m/s.
- The final deceleration segment: it goes from 1 square tall (27 m/s) to 0, over 1 square of time (1 s)? No, wait no—the last segment: from the end of the lower horizontal line to rest: the lower horizontal line is 2 squares tall? Wait no, let's do this properly:

Wait, the problem states that the time spent at maximum speed (189 m/s) is 3 seconds. On the graph, the horizontal line at maximum speed is 3 grid units long. So 1 grid unit on the t-axis = 1 second.

Maximum speed is 189 m/s, which is 7 grid units on the v-axis. So 1 grid unit on v-axis = $\frac{189}{7}=27$ m/s.

Now, the segment where the object comes to rest is the final downward line: it goes from a speed of 1 grid unit (27 m/s) to 0, over 1 grid unit of time (1 second)? No, wait no—the final segment: let's look at the graph: the last segment is a line from (let's say) $(t, v)$ to $(t+\Delta t, 0)$. The $\Delta t$ is 1 grid (1 s), and the change in velocity is $0 - 27 = -27$ m/s? No, wait no—wait, no, the lower horizontal segment is 2 grid units tall? Wait, no, the top horizontal is 7 grids (189), so each grid is 27. The lower horizontal is 2 grids: $2 \times 27=54$ m/s. Then the final deceleration is from 54 m/s to 0 over 1 grid (1 s)? No, no, the problem says "when it comes to rest", so the acceleration is the slope of the last segment:

Slope = $\frac{\Delta v}{\Delta t} = \frac{0 - v_{final\_constant}}{\Delta t}$

Wait, no, let's re-express:
We know that the time for maximum speed is 3 s, which is 3 grid squares, so 1 grid square on t-axis = 1 s.
Maximum speed is 189 m/s, which is 7 grid squares on v-axis, so 1 grid square on v-axis = $\frac{189}{7}=27$ m/s.

The segment where the object decelerates to rest: it goes from 1 grid square (27 m/s) to 0 over 1 grid square (1 s)? No, wait, no—the last segment is from the end of the lower horizontal line (which is 2 grid squares tall: $2 \times 27=54$ m/s) to 0, over 2 grid squares (2 s)? Wait, no, looking at the graph: the final downward line has a run of 2 squares (2 s) and a rise of -2 squares ($-2 \times 27 = -54$ m/s). No, wait, no—the top downward line: from 7 squares (189) to 2 squares (54) over 5 squares (5 s): slope $\frac{54-189}{5} = \frac{-135}{5}=-27$ m/s². But the final deceleration is from 2 squares (54) to 0 over 2 squares (2 s): slope $\frac{0-54}{2}=-27$? No, but the problem says "the moment when it comes to rest", which is the final deceleration.

Wait, no, let's use the total distance. Wait, no, the problem says "starting from rest, reaches 189 m/s, spends 3 s at that speed, then comes to rest". So total distance is:
Area under graph = area of first triangle + area of rectangle + area of last triangle.
Let $t_1$ = time to accelerate to 189, $t_3$ = time to decelerate to rest.
Total distance $d = \frac{1}{2} \times 189 \times t_1 + 189 \times 3 + \frac{1}{2} \times 189 \times t_3$
But from the graph, the acceleration segment (first line) has a run of 2 squares (2 s) to reach 2 squares (54 m/s)? No, no, I messed up. Wait, no—the graph:
- First segment: from (0,0) to (2, 2) grid units: speed goes from 0 to 54 m/s in 2 s: acceleration $a_1=\frac{54}{2}=27$ m/s².
- Second segment: horizontal from (2,2) to (5,2) grid units: 3 s (5-2=3), speed 54 m/s? No, but the problem says maximum speed is 189. Oh! I had the vertical axis backwards: the top of the graph is the maximum speed. So the vertical axis goes up, so the top horizontal line is the maximum speed, which is 7 grid units tall. The time for that is 3 grid units (3 s). So first segment: from (0,0) to (3,7): acceleration to 189 m/s in 3 s? No, 3 s to reach 189: $a=\frac{189}{3}=63$ m/s². Then horizontal from (3,7) to (6,7): 3 s (6-3=3) at 189 m/s. Then decelerates to (11,2): 11-6=5 s, speed goes from 189 to $2 \times \frac{189}{7}=54$ m/s: acceleration $a=\frac{54-189}{5}=\frac{-135}{5}=-27$ m/s². Then horizontal from (11,2) to (13,2): 2 s at 54 m/s. Then decelerates to (14,0): 14-13=1 s, speed from 54 to 0: acceleration $a=\frac{0-54}{1}=-54$ m/s². But the problem says "the object spends three seconds moving at its maximum speed of 189 meters per second"—that is the horizontal segment from (3,7) to (6,7), which is 3 grid units, so 3 s, correct. Then "when it comes to rest" is the last segment: from (13,2) to (14,0). So $\Delta v = 0 - 54 = -54$ m/s, $\Delta t = 14-13=1$ s. So acceleration $a=\frac{\Delta v}{\Delta t}=\frac{-54}{1}=-54$ m/s². But wait, no, the problem says "starting from rest at time t=0. The object spends three seconds moving at its maximum speed of 189 meters per second." So the maximum speed is 189, which is 7 grid units, so each grid is 27 m/s. The last segment is from 1 grid unit (27) to 0 over 1 s? No, I'm confused. Wait, let's use the definition: acceleration is $\frac{\Delta v}{\Delta t}$.

Wait, let's do this correctly:
1. Max speed $v_{max}=189$ m/s, time at max speed $t_2=3$ s.
2. On the v-t graph, the horizontal line for max speed is 3 grid squares, so 1 grid square on t-axis = 1 s.
3. The max speed is 7 grid squares on v-axis, so 1 grid square on v-axis = $\frac{189}{7}=27$ m/s.
4. The segment where the object comes to rest is the final downward line: it goes from 2 grid squares (2×27=54 m/s) to 0 over 1 grid square (1 s).
5. Acceleration $a = \frac{\Delta v}{\Delta t} = \frac{0 - 54}{1} = -54$ m/s². The negative sign means deceleration, which is acceleration opposite to motion.

Wait, but another way: the area under the graph is total distance. Let's calculate total distance:
- First triangle: from (0,0) to (2,7): area $\frac{1}{2}×2×189=189$ m
- Horizontal segment: from (2,7) to (5,7): area $3×189=567$ m
- Second triangle: from (5,7) to (10,2): area $\frac{1}{2}×(189+54)×5=\frac{1}{2}×243×5=607.5$ m
- Horizontal segment: from (10,2) to (12,2): area $2×54=108$ m
- Final triangle: from (12,2) to (13,0): area $\frac{1}{2}×1×54=27$ m
Total distance: $189+567+607.5+108+27=1498.5$ m, but we don't need that. The acceleration when coming to rest is the slope of the final segment: $\frac{0-54}{13-12}=-54$ m/s².

Wait, but maybe the final segment is from 1 grid square to 0: no, the max speed is 7 squares, so 2 squares is 54, which is the lower horizontal. The final segment is from 2 squares to 0 over 1 square, so $\Delta v=-54$, $\Delta t=1$, so $a=-54$ m/s².

Wait, but let's check the problem again: "Find the acceleration (in $\frac{m}{s^2}$) of the object at the moment when it comes to rest during this interval. [provide at least one decimal place]"

So the acceleration is -54.0 m/s² (or 54.0 m/s² deceleration, but acceleration is negative as it's opposite to velocity).

Wait, no, maybe I messed up the grid count. Let's count again:
- The top horizontal line (max speed) is 3 grid spaces long, which is 3 s, so 1 grid = 1 s.
- The top horizontal line is 7 grid spaces up from t-axis, so 7 grids = 189 m/s, so 1 grid = 27 m/s.
- The final downward line: starts at 2 grids above t-axis (2×27=54 m/s) and ends at 0, over 1 grid (1 s). So $\Delta v = 0 - 54 = -54$ m/s, $\Delta t=1$ s. Acceleration $a=\frac{\Delta v}{\Delta t}=-54.0$ m/s².

Yes, that makes sense.

Step1: Define grid scale for axes

1 grid (t-axis) = 1 s, 1 grid (v-axis) = $\frac{189}{7}=27$ m/s

Step2: Get final deceleration values

$\Delta v = 0 - (2 \times 27) = -54$ m/s, $\Delta t=1$ s

Step3: Calculate acceleration

$a = \frac{\Delta v}{\Delta t}$
$a = \frac{-54}{1} = -54.0$ m/s²

Answer:

$\boldsymbol{-54.0}$ $\frac{m}{s^2}$ (or magnitude 54.0 $\frac{m}{s^2}$ deceleration)