A cyclist, starting from rest, travels in a straight line for 10 minutes.
During the first 2.0 min of her trip, she maintains a uniform acceleration of 0.067 m/s2.
She then travels at constant velocity for the next 5.0 min.
Next, she decelerates at a constant rate so that she comes to rest 3.0 min later.
Create a graph of the cyclist's velocity versus time that closely approximates her trip. The individual events of the trip are represented by the dots at the bottom of the graph. Move each dot to the appropriate velocity.
What is the acceleration πlast during the last 3 min?
How far π· does the cyclist travel?
To find the acceleration πlast during the last 3 minutes, we need to calculate the change in velocity and divide it by the time interval.
Given:
Time interval: 3 minutes = 180 seconds
Acceleration during the first part of the trip: 0.067 m/s^2
Constant velocity duration: 5 minutes = 300 seconds
First, let's calculate the change in velocity during the first part of the trip.
Using the equation of motion:
π£ = π’ + ππ‘
Where:
π’ = initial velocity,
π£ = final velocity,
π = acceleration,
π‘ = time.
We start from rest, so π’ = 0 m/s.
Plugging in the values, we get:
π£ = 0 + (0.067 m/s^2) Γ (2.0 min Γ 60 s/min)
π£ = 0 + (0.067 m/s^2) Γ (120 s)
π£ = 0 + 8.04 m/s
π£ β 8.04 m/s
During the constant velocity phase, the velocity remains the same, so π£ = 8.04 m/s.
Now, let's calculate the change in velocity during the deceleration phase.
Again, using the equation of motion:
π£ = π’ + ππ‘
We want to find the final velocity π£ when the cyclist comes to rest, so π£ = 0 m/s.
We also know the time interval π‘ = 3 min = 180 s.
Plugging in the values and rearranging the equation, we get:
0 = 8.04 m/s + πlast Γ 180 s
Now we can solve for πlast:
πlast Γ 180 s = -8.04 m/s
πlast = -8.04 m/s Γ· 180 s
πlast β -0.0447 m/s^2
Thus, the acceleration πlast during the last 3 minutes is approximately -0.0447 m/s^2.
To find the distance D the cyclist travels, we need to calculate the area under the velocity-time graph.
First, let's calculate the area for the first phase of uniform acceleration.
The shape under this portion of the graph is a trapezoid.
The formula for the area of a trapezoid is:
Area = ((a+b)/2) Γ h
Where:
a = length of the first parallel side,
b = length of the second parallel side,
h = height (width) of the trapezoid.
In this case:
a = 0 m/s (initial velocity)
b = 8.04 m/s (final velocity)
h = 2.0 min Γ 60 s/min = 120 s (time duration)
Plugging in the values, we get:
Area1 = ((0 m/s + 8.04 m/s)/2) Γ 120 s
Area1 = (8.04 m/s / 2) Γ 120 s
Area1 β 482.4 m
Next, let's calculate the area for the constant velocity phase.
The shape under this portion of the graph is a rectangle. The formula for the area of a rectangle is:
Area = length Γ width
In this case, the width is the time duration during the constant velocity phase:
Time duration = 5.0 min Γ 60 s/min = 300 s
Plugging in the values, we get:
Area2 = 8.04 m/s Γ 300 s
Area2 = 2412 m
Lastly, let's calculate the area for the deceleration phase.
The shape under this portion of the graph is another trapezoid.
Using the same formula for the area of a trapezoid, we find:
Area3 = ((8.04 m/s + 0 m/s)/2) Γ 180 s
Area3 = 4.02 m/s Γ 180 s
Area3 = 723.6 m
To find the total distance, we sum up the three areas:
Total distance = Area1 + Area2 + Area3
Total distance = 482.4 m + 2412 m + 723.6 m
Total distance β 3618 m
Thus, the cyclist travels approximately 3618 meters (or 3.618 kilometers).