Consider the velocity vs time graph of a person in an elevator shown in. Suppose the elevator is initially at rest. It than accelerates from 3 seconds maintain that velocity for 15 seconds, than decelerate to 5 seconds from motion equation for constant acceleration in one dimension for the complete trip. (We could however, use them in three individual section where acceleration is constant.) Sketch graphs of (a) position vs time (b) acceleration vs time for the trip
To sketch the position vs time graph, we need to integrate the velocity vs time graph. We can break down the trip into three sections: the acceleration phase, the constant velocity phase, and the deceleration phase.
(a) Position vs time graph:
1. Acceleration phase (from 0 s to 3 s):
Since the elevator starts from rest and accelerates, the velocity vs time graph will be a straight line with a positive slope. Integrating this part will give us a parabolic curve for the position vs time graph.
2. Constant velocity phase (from 3 s to 18 s):
During this phase, the elevator maintains a constant velocity, which means the velocity vs time graph will be a horizontal line. Integrating a horizontal line will give us a straight line in the position vs time graph.
3. Deceleration phase (from 18 s to 23 s):
The elevator decelerates, so the velocity vs time graph will be a straight line with a negative slope. Integrating this part will give us a parabolic curve for the position vs time graph.
Overall, the position vs time graph will look like this:
```
| __________
| / \
| / \
| / \
| / \
______|/_________________\_____
0s 3s 18s 23s
```
(b) Acceleration vs time graph:
From the given information, we know the elevator accelerates for 3 seconds, maintains a constant velocity for 15 seconds, and then decelerates for 5 seconds.
The acceleration vs time graph will look like this:
```
/|
/ |
/ |
/ |
/____|\
0s 3s 18s 23s
```
The acceleration is initially positive during the acceleration phase, then becomes zero during the constant velocity phase, and finally becomes negative during the deceleration phase.
To sketch the graphs of position vs time and acceleration vs time for the given scenario, we will need to analyze the different phases of the elevator's motion.
Phase 1: Initial Rest
In this phase, the elevator is at rest, so the velocity is zero. The acceleration is also zero because there is no change in velocity. The position remains constant at the starting point.
Phase 2: Acceleration
During this phase, the elevator accelerates for 3 seconds. Let's assume the acceleration is constant at a value of 'a'.
We know that the velocity-time relationship for constant acceleration is given by the equation:
v = u + at
Since the elevator starts from rest (initial velocity u = 0), the equation simplifies to:
v = at
The position-time relationship for constant acceleration is given by the equation:
s = ut + (1/2)at^2
Since the initial velocity is zero, the equation further simplifies to:
s = (1/2)at^2
Phase 3: Constant Velocity
After the acceleration phase, the elevator maintains a constant velocity for 15 seconds. This means there is no acceleration during this period, and the velocity remains constant at 'v' achieved at the end of the acceleration phase. The position of the elevator increases linearly with time during this phase because there is no acceleration.
Phase 4: Deceleration
In the final phase, the elevator decelerates for 5 seconds. The acceleration during this phase is constant and in the opposite direction. Let's assume the acceleration is 'b' (negative value).
To summarize:
Phase 1: Initial rest (0-3 seconds)
- Velocity = 0 m/s
- Acceleration = 0 m/s^2
- Position remains constant
Phase 2: Acceleration (3-18 seconds)
- Velocity = a*t (where 't' is the time in seconds)
- Position = (1/2)*a*t^2 (where 't' is the time in seconds)
Phase 3: Constant velocity (18-33 seconds)
- Velocity remains constant (v)
- Position increases linearly with time
Phase 4: Deceleration (33-38 seconds)
- Velocity = v + b*t (where 't' is the time in seconds)
- Position = (v*t) + (1/2)*b*t^2 (where 't' is the time in seconds)
Now, using the above information, you can sketch the position vs time graph and acceleration vs time graph for the complete trip.
To sketch the position vs time and acceleration vs time graphs for the given scenario, we need to understand the motion equations for constant acceleration in one dimension. These equations are:
1. Position equation:
- s = ut + (1/2)at²
where s represents the position, u is the initial velocity, t is the time, and a is the acceleration.
2. Velocity equation:
- v = u + at
where v represents the final velocity.
3. Acceleration equation:
- a = (v - u) / t
Now let's break down the elevator's motion into its three individual sections:
Section 1: Acceleration from rest (0 m/s) to final velocity
- Initial velocity (u1) = 0 m/s
- Final velocity (v1) = ? (we will calculate it)
- Time (t1) = 3 seconds
- Acceleration (a1) = v1 / t1
Section 2: Constant velocity
- Initial velocity (u2) = v1 (final velocity from the previous section)
- Final velocity (v2) = v1 (same as initial velocity)
- Time (t2) = 15 seconds
- Acceleration (a2) = 0 m/s² (constant velocity, so no acceleration)
Section 3: Deceleration to rest (0 m/s)
- Initial velocity (u3) = v1 (final velocity from the previous section)
- Final velocity (v3) = 0 m/s
- Time (t3) = 5 seconds
- Acceleration (a3) = (v3 - u3) / t3
Now let's calculate the values for each section.
Section 1:
- Taking u1 = 0 m/s, t1 = 3 seconds:
a1 = v1 / t1
Let's assume v1 = final velocity = 10 m/s (for example)
Therefore, a1 = 10 m/s / 3 s ≈ 3.33 m/s²
Section 2:
- Velocity remains constant at 10 m/s (from the previous section)
Section 3:
- Taking t3 = 5 seconds, v3 = 0 m/s, and u3 = 10 m/s:
a3 = (v3 - u3) / t3
a3 = (0 - 10 m/s) / 5 s = -2 m/s²
Now, we can sketch the graphs:
(a) Position vs Time graph:
The position vs time graph is obtained by integrating the velocity equation.
Here, the initial position is assumed to be 0.
- For Section 1, using the position equation: s1 = (1/2)a1t1²
s1 = (1/2)(3.33 m/s²)(3 s)² = 15 m
- For Section 2, since velocity is constant, position increases linearly:
s2 = v2 * t2 = (10 m/s)(15 s) = 150 m
- For Section 3, using the position equation: s3 = (v3 - u3) / t3 * t3² + u3 * t3
s3 = (-2 m/s²)(5 s)² + (10 m/s)(5 s) = 25 m
So, the position vs time graph would show a linear increase in position during Section 2, with discontinuous segments at the beginning and end.
(b) Acceleration vs Time graph:
The acceleration vs time graph shows the instantaneous acceleration at each point in time.
- In Section 1, we calculated a1 = 3.33 m/s².
- In Section 2, there is no acceleration, so a2 = 0 m/s².
- In Section 3, we calculated a3 = -2 m/s².
So, the acceleration vs time graph would show a positive constant acceleration in Section 1, no acceleration in Section 2, and a negative constant acceleration in Section 3.
By following these calculations and using the motion equations for constant acceleration, you can sketch the position vs time and acceleration vs time graphs for the elevator's trip.