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!