Calculate (i) Acceleration from O to A (ii) Retardation State the distance covered by Mr Hara while moving at (i) constant speed (ii) Acceleration What was the distance covered between point B and C

Here are the calculations based on the provided speed-time graph:

1. Calculate

• (i) Acceleration from O to A Acceleration is the gradient of the speed-time graph. Step 1: Identify the initial and final speeds and times for segment OA. At point O: initial speed $v_i = 0$ m/s, initial time $t_i = 0$ s. At point A: final speed $v_f = 20$ m/s, final time $t_f = 6$ s. Step 2: Apply the acceleration formula. $a = \frac{v_f - v_i}{t_f - t_i}$ $a = \frac{20\, m/s - 0\, m/s}{6\, s - 0\, s}$ $a = \frac{20}{6}\, m/s^2$ $a = \frac{10}{3}\, m/s^2$ $a \approx 3.33\, m/s^2$ The acceleration from O to A is $\boxed{3.33 m/s^2}$.

• (ii) Retardation Retardation (or deceleration) occurs when the speed decreases. Step 1: Examine the graph for any segments where the speed is decreasing. The graph shows segments of increasing speed (OA, BC, DE) and constant speed (AB, CD). There are no segments where the speed decreases. There is no retardation shown in the given graph.

2. State the distance covered by Mr. Hara while moving at

The distance covered is the area under the speed-time graph.

• (i) Constant speed Constant speed occurs in segments AB and CD. Step 1: Calculate the distance for segment AB (rectangle). Speed = 20 m/s, Time = $10\, s - 6\, s = 4\, \text{s}$. Distance$_{AB} = speed \times time = 20\, m/s \times 4\, s = 80\, \text{m}$. Step 2: Calculate the distance for segment CD (rectangle). Speed = 40 m/s, Time = $18\, s - 14\, s = 4\, \text{s}$. Distance$_{CD} = speed \times time = 40\, m/s \times 4\, s = 160\, \text{m}$. Step 3: Sum the distances. Total distance at constant speed = Distance$_{AB}$ + Distance$_{CD} = 80\, m + 160\, m = 240\, \text{m}$. The distance covered at constant speed is $\boxed{240 m}$.

• (ii) Acceleration Acceleration occurs in segments OA, BC, and DE. Step 1: Calculate the distance for segment OA (triangle). Base = 6 s, Height = 20 m/s. Distance$_{OA} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 6\, s \times 20\, m/s = 60\, \text{m}$. Step 2: Calculate the distance for segment BC (trapezium). Parallel sides $v_1 = 20$ m/s, $v_2 = 40$ m/s. Height (time) = $14\, s - 10\, s = 4\, \text{s}$. Distance$_{BC} = \frac{1}{2} \times (v_1 + v_2) \times time = \frac{1}{2} \times (20\, m/s + 40\, m/s) \times 4\, s = \frac{1}{2} \times 60\, m/s \times 4\, s = 120\, \text{m}$. Step 3: Calculate the distance for segment DE (trapezium). Parallel sides $v_1 = 40$ m/s, $v_2 = 80$ m/s. Height (time) = $30\, s - 18\, s = 12\, \text{s}$. Distance$_{DE} = \frac{1}{2} \times (v_1 + v_2) \times time = \frac{1}{2} \times (40\, m/s + 80\, m/s) \times 12\, s = \frac{1}{2} \times 120\, m/s \times 12\, s = 720\, \text{m}$. Step 4: Sum the distances. Total distance during acceleration = Distance$_{OA}$ + Distance$_{BC}$ + Distance$_{DE} = 60\, m + 120\, m + 720\, m = 900\, \text{m}$. The distance covered during acceleration is $\boxed{900 m}$.

3. What was the distance covered between point B and C?

This refers to the distance calculated for segment BC in the previous question. Step 1: Identify the speeds and times at points B and C. At point B: speed $v_B = 20$ m/s, time $t_B = 10$ s. At point C: speed $v_C = 40$ m/s, time $t_C = 14$ s. Step 2: Calculate the area of the trapezium formed by segment BC. $\text{Distance}_{BC} = \frac{1}{2} \times (v_B + v_C) \times (t_C - t_B)$ $\text{Distance}_{BC} = \frac{1}{2} \times (20\, m/s + 40\, m/s) \times (14\, s - 10\, s)$ $\text{Distance}_{BC} = \frac{1}{2} \times 60\, m/s \times 4\, \text{s}$ $\text{Distance}_{BC} = 120\, \text{m}$ The distance covered between point B and C is $\boxed{120 m}$.

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