The diagram shows the speed time graph of a moving object. a) Find the acceleration of the object in the first 3 seconds. b) Given that the object covered a distance of 115m in the first 10 seconds find the value of v. c) calculate the average speed of the object in the first 14 seconds

Here's how to solve the problem using the provided speed-time graph:

a) Find the acceleration of the object in the first 8 seconds.

Step 1: Identify the initial and final speeds and times. From the graph, at $t=0$ s, the initial speed is $u=0$ m/s. At $t=8$ s, the final speed is $v_f=20$ m/s. Acceleration is the gradient of the speed-time graph. $a = \frac{\text{change in speed}}{\text{change in time}}$ $a = \frac{v_f - u}{\Delta t}$ $a = \frac{20 \text{ m/s} - 0 \text{ m/s}}{8 \text{ s} - 0 \text{ s}}$ $a = \frac{20}{8} \text{ m/s}^2$ $a = \boxed{\text{*2.5 m/s}^2*}$

b) Given that the object covered a distance of 115m in the first 10 seconds, find the value of v.

Step 2: Calculate the area under the graph for the first 10 seconds. The distance covered is the area under the speed-time graph. The area from $t=0$ s to $t=10$ s consists of a triangle (0-8s) and a trapezium (8-10s). Distance for the first 8 seconds (triangle): $D_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \text{ s} \times 20 \text{ m/s} = 80 \text{ m}$ Distance from 8 s to 10 s (trapezium): At $t=8$ s, speed is $20$ m/s. At $t=10$ s, speed is $v$ m/s. $D_2 = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$ $D_2 = \frac{1}{2} \times (20 \text{ m/s} + v \text{ m/s}) \times (10 \text{ s} - 8 \text{ s})$ $D_2 = \frac{1}{2} \times (20 + v) \times 2 = (20 + v) \text{ m}$ The total distance in the first 10 seconds is $D_1 + D_2$. $115 \text{ m} = 80 \text{ m} + (20 + v) \text{ m}$ $115 = 100 + v$ $v = 115 - 100$ $v = \boxed{\text{*15 m/s*}}$

c) Calculate the average speed of the object in the first 14 seconds.

Step 3: Calculate the total distance in the first 14 seconds and then the average speed. Average speed is total distance divided by total time. Total time is $14$ s. The total distance in the first 14 seconds is the sum of the areas: triangle (0-8s), trapezium (8-10s), and rectangle (10-14s). We know $D_1 = 80$ m. Using $v=15$ m/s from part b, $D_2 = (20 + 15) \text{ m} = 35 \text{ m}$. Distance from 10 s to 14 s (rectangle): The speed is constant at $v=15$ m/s from $t=10$ s to $t=14$ s. $D_3 = \text{base} \times \text{height} = (14 \text{ s} - 10 \text{ s}) \times 15 \text{ m/s}$ $D_3 = 4 \text{ s} \times 15 \text{ m/s} = 60 \text{ m}$ Total distance in the first 14 seconds: $D_{\text{total}} = D_1 + D_2 + D_3 = 80 \text{ m} + 35 \text{ m} + 60 \text{ m} = 175 \text{ m}$ Now, calculate the average speed: $\text{Average speed} = \frac{D_{\text{total}}}{\text{Total time}}$ $\text{Average speed} = \frac{175 \text{ m}}{14 \text{ s}}$ $\text{Average speed} = \boxed{\text{*12.5 m/s*}}$