Question

a bus moves from rest with uniform acceleration of 2m/s² for the first 10 secs. It then accelerate at uniform rate of 1m/s² for another 15 secs. It continues at a constant speed for 70 secs and finally comes to rest in 20 secs by uniform retardation. draw the velocity time graph and: 1. Find the total distance traveled 2. Average speed for the journey 3. the retardation of the journey 4. the maximum speed attained

Answer 1

Let's analyze the bus's journey step by step and create a velocity-time graph.

### Breakdown of the Journey

1. **First Segment: Accelerating from Rest**
- **Initial Velocity (u)** = 0 m/s
- **Acceleration (a)** = \(2 \, \text{m/s}^2\)
- **Time (t)** = 10 s

\[
v = u + at = 0 + (2)(10) = 20 \, \text{m/s}
\]

2. **Second Segment: Accelerating**
- **Initial Velocity (u)** = 20 m/s (from the last segment)
- **Acceleration (a)** = \(1 \, \text{m/s}^2\)
- **Time (t)** = 15 s

\[
v = u + at = 20 + (1)(15) = 35 \, \text{m/s}
\]

3. **Third Segment: Constant Speed**
- **Velocity** = \(35 \, \text{m/s}\)
- **Time (t)** = 70 s

4. **Fourth Segment: Decelerating to Rest**
- **Initial Velocity (u)** = 35 m/s
- **Final Velocity (v)** = 0 m/s
- **Time (t)** = 20 s

Let's use the formula \(v = u + at\) to find the retardation (negative acceleration, \(a\)):
\[
0 = 35 + a (20)
\]
\[
a = -\frac{35}{20} = -1.75 \, \text{m/s}^2
\]

### Velocity-Time Graph

1. **Segment 1**: From (0, 0) to (10, 20)
2. **Segment 2**: From (10, 20) to (25, 35)
3. **Segment 3**: From (25, 35) to (95, 35)
4. **Segment 4**: From (95, 35) to (115, 0)

### 1. Total Distance Traveled

- **First Segment:**
\[
d_1 = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} (2)(10^2) = 100 \, \text{m}
\]

- **Second Segment:**
Using \(d = ut + \frac{1}{2} a t^2\), where \(u = 20 \, \text{m/s}\), \(a = 1 \, \text{m/s}^2\), and \(t = 15 \, \text{s}\):

\[
d_2 = (20)(15) + \frac{1}{2}(1)(15^2) = 300 + 112.5 = 412.5 \, \text{m}
\]

- **Third Segment:**
\[
d_3 = \text{velocity} \times \text{time} = 35 \cdot 70 = 2450 \, \text{m}
\]

- **Fourth Segment:**
Using the formula \(d = ut + \frac{1}{2} a t^2\) implies:
\[
d_4 = (35)(20) + \frac{1}{2}(-1.75)(20^2) = 700 - 350 = 350 \, \text{m}
\]

- **Total Distance**
\[
\text{Total Distance} = d_1 + d_2 + d_3 + d_4 = 100 + 412.5 + 2450 + 350 = 3312.5 \, \text{m}
\]

### 2. Average Speed for the Journey

- **Total Time**: \(10 + 15 + 70 + 20 = 115 \, \text{s}\)

- **Average Speed**:
\[
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{3312.5 \, \text{m}}{115 \, \text{s}} \approx 28.84 \, \text{m/s}
\]

### 3. Retardation of the Journey
The retardation of the journey is \(1.75 \, \text{m/s}^2\).

### 4. Maximum Speed Attained
The maximum speed attained is \(35 \, \text{m/s}\).

### Summary:
1. Total Distance Traveled: **3312.5 m**
2. Average Speed for the Journey: **28.84 m/s**
3. Retardation of the Journey: **1.75 m/s²**
4. Maximum Speed Attained: **35 m/s**

Please let me know if you would like further assistance or clarification on any aspect!