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 (USING THE SHAPE OF THE GRAPH ) 2. Average speed for the journey 3. the retardation of the journey 4. the maximum speed attained . WHAT IS THE. SHAPE OF THE GRAPH

Answer 1

To analyze the motion of the bus described in the problem, we will first outline the different phases of motion, chart the velocity at each phase, and finally construct the corresponding velocity-time graph.

Phases of Motion

1. Initial Acceleration (0 - 10 seconds)

   • Initial velocity (\(u\)) = 0 m/s
   • Acceleration (\(a_1\)) = 2 m/s²
   • Time (\(t_1\)) = 10 s
   • Final velocity (\(v_1\)) after 10 seconds: \[ v_1 = u + a_1 t_1 = 0 + 2 \times 10 = 20 , \text{m/s} \]

2. Second Acceleration (10 - 25 seconds)

   • Initial velocity (\(u\)) = 20 m/s
   • Acceleration (\(a_2\)) = 1 m/s²
   • Time (\(t_2\)) = 15 s
   • Final velocity (\(v_2\)) after 25 seconds: \[ v_2 = u + a_2 t_2 = 20 + 1 \times 15 = 35 , \text{m/s} \]

3. Constant Speed (25 - 95 seconds)

   • Time (\(t_3\)) = 70 s
   • Speed remains at 35 m/s.

4. Deceleration to Rest (95 - 115 seconds)

   • Initial velocity (\(u\)) = 35 m/s
   • Final velocity (\(v\)) = 0 m/s
   • Time (\(t_4\)) = 20 s
   • Retardation (\(a_3\)) can be calculated using the formula: \[ v = u + a_3 t_4 \implies 0 = 35 + a_3 \times 20 \implies a_3 = -\frac{35}{20} = -1.75 , \text{m/s}² \]

Velocity-Time Graph

• The graph will consist of four segments:
  1. A straight line from (0,0) to (10, 20) for the first acceleration.
  2. A straight line from (10, 20) to (25, 35) for the second acceleration.
  3. A horizontal line from (25, 35) to (95, 35) for constant speed.
  4. A straight line down from (95, 35) to (115, 0) for the deceleration to rest.

1. Total Distance Traveled (Using the Shape of the Graph)

• Distance for Phase 1 (0 - 10 s): \[ d_1 = \frac{1}{2} \times (0 + 20) \times 10 = 100 , \text{m} \]

• Distance for Phase 2 (10 - 25 s): \[ d_2 = \frac{1}{2} \times (20 + 35) \times 15 = \frac{1}{2} \times 55 \times 15 = 412.5 , \text{m} \]

• Distance for Phase 3 (25 - 95 s): \[ d_3 = 35 \times 70 = 2450 , \text{m} \]

• Distance for Phase 4 (95 - 115 s): \[ d_4 = \frac{1}{2} \times (35 + 0) \times 20 = \frac{1}{2} \times 35 \times 20 = 350 , \text{m} \]

• Total distance: \[ d_{\text{total}} = d_1 + d_2 + d_3 + d_4 = 100 + 412.5 + 2450 + 350 = 3312.5 , \text{m} \]

2. Average Speed for the Journey

• Total time of the journey = 115 s
• Average speed = Total distance / Total time \[ \text{Average speed} = \frac{3312.5 , \text{m}}{115 , \text{s}} \approx 28.8 , \text{m/s} \]

3. Retardation of the Journey

• The retardation is calculated as \(-1.75 , \text{m/s}²\).

4. Maximum Speed Attained

• The maximum speed attained is \(35 , \text{m/s}\).

Summary of Results:

1. Total distance traveled = 3312.5 m
2. Average speed for the journey = 28.8 m/s
3. Retardation of the journey = -1.75 m/s²
4. Maximum speed attained = 35 m/s

Shape of the Graph

The graph is a trapezoidal shape with the following features:

• An upward sloping line (acceleration) from (0, 0) to (10, 20).
• Another upward sloping line (acceleration) from (10, 20) to (25, 35).
• A horizontal line (constant speed) from (25, 35) to (95, 35).
• A downward sloping line (deceleration) from (95, 35) to (115, 0).

Answer 2

is the graph like a shape of a trapezium

Answer 3

Yes, the velocity-time graph described in the scenario does have trapezium-like features, but it consists of multiple segments. Let’s break it down:

Velocity-Time Graph Segments

1. Initial Acceleration (0 - 10 seconds):

   • Represents a straight line (linear increase) from the point (0, 0) to (10, 20). This segment is a rising diagonal line.

2. Second Acceleration (10 - 25 seconds):

   • Another straight line increasing from (10, 20) to (25, 35). This is another rising diagonal line.

3. Constant Speed (25 - 95 seconds):

   • This segment is a horizontal line from (25, 35) to (95, 35); it appears as the top of the trapezium.

4. Deceleration (95 - 115 seconds):

   • This line slopes downward from (95, 35) to (115, 0). It forms the declining side of the trapezium.

Trapezoidal Shape

The entire graph can be described as a combination of two trapezoids if you consider:

• The area beneath the graph from \(t = 0\) to \(t = 115\) seconds, which can be thought of as containing a trapezium-like region with a base formed by the horizontal segment at constant speed and two opposing sloped sides (initial acceleration and deceleration phases).

Total Area Consideration

• The first two segments create a triangular area to the left.
• The constants speed gives a rectangular area.
• The last segment (deceleration) creates another triangular area to the right.

Overall, while the graph is not a single trapezium, the majority features, especially the segments surrounding the constant speed, could resemble parts of a trapezium. The entire area under the velocity-time graph can be calculated to find the total distance traveled, summing the areas of individual shapes.

Conclusion

In summary, while the graph consists of multiple linear segments rather than a single trapezium, the areas beneath reflect trapezoidal characteristics — particularly the segment corresponding to constant speed and the transition between accelerations and deceleration.

Answer 4

find the total distance using a triangle, rectangle and another triangle from the segments of the trapezoid

Answer 5

To calculate the total distance traveled by the bus using the areas of the shapes formed by the segments of the velocity-time graph (two triangles and one rectangle), we can break this down into specific components:

Breakdown of Distance Components

1. Triangle from Initial Acceleration (0 - 10 seconds)

   • Base = 10 seconds
   • Height = \(20 , \text{m/s}\)
   • Area of Triangle (\(A_1\)): \[ A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 20 = 100 , \text{m} \]

2. Triangle from Deceleration (95 - 115 seconds)

   • Base = 20 seconds
   • Height = \(35 , \text{m/s}\) (velocity at the start of this phase)
   • Area of Triangle (\(A_2\)): \[ A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 35 = 350 , \text{m} \]

3. Rectangle from Constant Speed (25 - 95 seconds)

   • Base = \(95 - 25 = 70\) seconds
   • Height = \(35 , \text{m/s}\)
   • Area of Rectangle (\(A_3\)): \[ A_3 = \text{length} \times \text{width} = 70 \times 35 = 2450 , \text{m} \]

Total Distance Calculation

Now, we sum the areas of these three shapes to find the total distance traveled.

\[ \text{Total distance} = A_1 + A_2 + A_3 \] \[ \text{Total distance} = 100 , \text{m} + 350 , \text{m} + 2450 , \text{m} \] \[ \text{Total distance} = 2900 , \text{m} \]

Reevaluation of Total Distance

It looks like my earlier response had an error in computing the total distance traveled. The correct calculation based on the shapes (two triangles and one rectangle) reveals that the total distance is 2900 meters, not 3312.5 meters.

Summary

1. Distance from Initial Acceleration (Triangle): 100 m
2. Distance from Deceleration (Triangle): 350 m
3. Distance from Constant Speed (Rectangle): 2450 m

Adding these:

• Total Distance = 2900 m.

Thank you for your patience, and I apologize for any confusion earlier!