Let's break down the motion of the bus and then use that to create a velocity-time (v-t) graph, which will help us calculate the total distance traveled.
### 1. Uniform Acceleration Phase:
- **Initial velocity (u)**: 0 m/s (starting from rest)
- **Acceleration (a)**: 0.6 m/s²
- **Time (t₁)**: 20 s
We can calculate the final velocity at the end of the acceleration phase using the equation:
\[ v = u + at \]
\[ v = 0 + (0.6 \, \text{m/s}^2 \times 20 \, \text{s}) \]
\[ v = 12 \, \text{m/s} \]
So, the bus reaches a maximum velocity of 12 m/s after 20 seconds.
### 2. Constant Velocity Phase:
- **Velocity (v)**: 12 m/s
- **Time (t₂)**: 40 s
### 3. Uniform Retardation Phase:
- **Final velocity (v)**: 0 m/s (comes to a stop)
- **Time (t₃)**: 10 s
We can calculate the retardation using the equation:
\[ v = u + at \]
\[ 0 = 12 + a(10) \]
\[ a = -\frac{12}{10} \, \text{m/s}^2 \]
\[ a = -1.2 \, \text{m/s}^2 \]
### Velocity-Time Graph:
The v-t graph will consist of three segments:
1. A straight line with positive slope representing the uniform acceleration from 0 to 12 m/s over 20 seconds.
2. A horizontal line representing the constant velocity of 12 m/s for the next 40 seconds.
3. A straight line with a negative slope representing the uniform retardation from 12 m/s to 0 m/s over the final 10 seconds.
### Calculating the Total Distance:
The area under the v-t graph represents the total distance traveled.
#### 1. Distance during uniform acceleration:
\[ \text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ = \frac{1}{2} \times 20 \, \text{s} \times 12 \, \text{m/s} \]
\[ = 120 \, \text{m} \]
#### 2. Distance during constant velocity:
\[ \text{Distance} = \text{velocity} \times \text{time} \]
\[ = 12 \, \text{m/s} \times 40 \, \text{s} \]
\[ = 480 \, \text{m} \]
#### 3. Distance during uniform retardation:
\[ \text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ = \frac{1}{2} \times 10 \, \text{s} \times 12 \, \text{m/s} \]
\[ = 60 \, \text{m} \]
### Total Distance:
\[ \text{Total Distance} = 120 \, \text{m} + 480 \, \text{m} + 60 \, \text{m} \]
\[ = 660 \, \text{m} \]
Therefore, the total distance between the two bus stops is 660 meters.
### Visual Representation:
Here’s how the velocity-time graph looks:
1. **From \(t = 0\) to \(t = 20\) seconds**: A linearly increasing line (slope = 0.6) from \(0 \, \text{m/s}\) to \(12 \, \text{m/s}\).
2. **From \(t = 20\) to \(t = 60\) seconds**: A horizontal line at \(12 \, \text{m/s}\).
3. **From \(t = 60\) to \(t = 70\) seconds**: A linearly decreasing line (slope = -1.2) from \(12 \, \text{m/s}\) to \(0 \, \text{m/s}\).
This graph accurately represents the three phases of the bus's journey: acceleration, constant velocity, and deceleration.
A passenger bus starts from a stop and travels with uniform acceleration of 0.6m/s^2 for 20s until it reached a maximum velocity.it travels with this maximum velocity for another 40seconds the brake is then applied so that a uniform retardation is obtained and then come to a halt at the next bus stop after 10s.sketch a velocity time graph of this motion and use this to calculate the total distance between the two bus stops.
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