In heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5 minutes, and finally you drive at 15 m/s for another 2.5 minutesPlot a position-versus-time graph for this motion. Your plot should extend from t=0 to t=7.5 minutes.
Assume x=0 and t=0 at the start of your motion.
Part B
Use your plot from part A to calculate the average velocity between t=0 to t=7.5 minutes.
7.4 m/s
To plot a position-versus-time graph for the given scenario, we can break it down into three different intervals:
Interval 1: Driving at 12 m/s for 1.5 minutes
Interval 2: Being stationary for 3.5 minutes
Interval 3: Driving at 15 m/s for 2.5 minutes
Since the position starts at x=0 and t=0, we can calculate the positions at different times for each interval.
Interval 1:
Position after t=0.025 minutes (1.5 seconds) = 12 m/s * 0.025 minutes = 0.3 meters
Position after t=0.05 minutes (3 seconds) = 12 m/s * 0.05 minutes = 0.6 meters
...
Position after t=1.5 minutes = 12 m/s * 1.5 minutes = 18 meters
Interval 2:
Since the car is stationary, the position remains constant at 18 meters for 3.5 minutes.
Interval 3:
Position after t=5 minutes = 18 meters + (15 m/s * 2.5 minutes) = 18 meters + 37.5 meters = 55.5 meters
Now, we can plot these positions on the graph:
```
Position (m)
|
| x
| | x
| x x |
| | | |
| |___x______|____|__________________
0 1.5 5 t (minutes)
```
Finally, to calculate the average velocity between t=0 and t=7.5 minutes, we need to find the total displacement and divide it by the total time. The displacement is the difference between the initial and final positions:
Displacement = (final position) - (initial position) = 55.5 meters - 0 meters = 55.5 meters
The total time is t=7.5 minutes - t=0 minutes = 7.5 minutes
Average velocity = Displacement / Total time = 55.5 meters / 7.5 minutes = 7.4 m/s
To plot a position-versus-time graph for this motion, we need to understand the different segments of the journey and calculate the position at each point in time.
First, let's convert the given time durations to seconds for easier calculations. 1.5 minutes, 3.5 minutes, and 2.5 minutes are equal to 90 seconds, 210 seconds, and 150 seconds respectively.
Now, let's break down the motion into three segments:
Segment 1: Driving at 12 m/s for 1.5 minutes (or 90 seconds).
During this segment, the position changes at a constant rate since the car is moving at a constant velocity. The formula to calculate the change in position is given by:
Δx = v * t
Here, v represents the velocity (12 m/s) and t represents the time (90 seconds).
Δx = 12 m/s * 90 s = 1080 m
So, during this segment, the car travels 1080 meters.
Segment 2: Stopping for 3.5 minutes (or 210 seconds).
Since the car stops, there is no change in position during this segment. Therefore, the position remains constant at 1080 meters.
Segment 3: Driving at 15 m/s for 2.5 minutes (or 150 seconds).
Similar to segment 1, the position changes at a constant rate because the car is moving at a constant velocity. Using the same formula as before:
Δx = v * t
Here, v represents the velocity (15 m/s) and t represents the time (150 seconds).
Δx = 15 m/s * 150 s = 2250 m
Therefore, during this segment, the car travels 2250 meters.
Now that we have the distances traveled during each segment, we can plot the position-versus-time graph:
At t = 0 minutes, the car is at the starting point, so the position is 0 meters.
Between t = 0 and t = 1.5 minutes (90 seconds), the car covers a distance of 1080 meters. Therefore, at t = 1.5 minutes, the position is 1080 meters.
At t = 1.5 minutes, the car stops for 3.5 minutes (210 seconds). Therefore, the position remains constant at 1080 meters until t = 5 minutes.
Between t = 5 minutes and t = 7.5 minutes (150 seconds), the car covers a distance of 2250 meters. Therefore, at t = 7.5 minutes, the position is 3330 meters.
The position-versus-time graph would be a straight line with a positive slope (indicating constant velocity) from t = 0 to t = 1.5 minutes, then a flat line at 1080 meters from t = 1.5 to t = 5 minutes, and finally another straight line with the same positive slope from t = 5 to t = 7.5 minutes.
Now, let's move on to part B and calculate the average velocity between t = 0 and t = 7.5 minutes.
Average velocity is calculated using the formula:
Average velocity = total displacement / total time
Total displacement in this case would be the final position minus the initial position, which is 3330 meters - 0 meters = 3330 meters.
Total time is 7.5 minutes or 450 seconds.
Average velocity = 3330 m / 450 s ≈ 7.4 m/s
Therefore, the average velocity between t = 0 and t = 7.5 minutes is approximately 7.4 m/s.