Question 3 [5 marks] Find the total distance travelled, the distance travelled in the first 40 seconds, the acceleration when t = 25.

Here's how to solve the problem using the provided velocity-time graph. a) The total distance travelled: The total distance travelled is the area under the velocity-time graph. We can divide the graph into three sections: a triangle, a rectangle, and a trapezium. Step 1: Calculate the area of the first section (triangle from t=0 to t=10 s). The base is 10 s and the height is 40 m/s. Area_1 = (1)/(2) × base × height = (1)/(2) × 10 s × 40 m/s = 200 m Step 2: Calculate the area of the second section (rectangle from t=10 to t=30 s). The width is 30 s - 10 s = 20 s and the height is 40 m/s. Area_2 = width × height = 20 s × 40 m/s = 800 m Step 3: Calculate the area of the third section (trapezium from t=30 to t=50 s). The parallel sides are 40 m/s (at t=30) and 80 m/s (at t=50). The height (time interval) is 50 s - 30 s = 20 s. Area_3 = (1)/(2) × (sum of parallel sides) × height = (1)/(2) × (40 m/s + 80 m/s) × 20 s = (1)/(2) × 120 m/s × 20 s = 1200 m Step 4: Calculate the total distance. Total Distance = Area_1 + Area_2 + Area_3 = 200 m + 800 m + 1200 m = 2200 m The total distance travelled is 2200 m. b) The distance travelled in the first 40 seconds: This is the area under the graph from t=0 to t=40 s. This includes the first two sections and part of the third section. Step 1: Determine the velocity at t=40 s. The acceleration between t=30 s and t=50 s is constant. a = ( v)/( t) = 80 m/s - 40 m/s50 s - 30 s = 40 m/s20 s = 2 m/s^2 Using v = u + at', where u=40 m/s at t'=0 (which corresponds to t=30 s), and t'=10 s (which corresponds to t=40 s): v_40 = 40 m/s + (2 m/s^2 × 10 s) = 40 m/s + 20 m/s = 60 m/s Step 2: Calculate the area of the third part up to t=40 s (trapezium from t=30 to t=40 s). The parallel sides are 40 m/s (at t=30) and 60 m/s (at t=40). The height (time interval) is 40 s - 30 s = 10 s. Area_30-40 = (1)/(2) × (40 m/s + 60 m/s) × 10 s = (1)/(2) × 100 m/s × 10 s = 500 m Step 3: Calculate the total distance in the first 40 seconds. Distance_0-40 = Area_1 + Area_2 + Area_30-40 = 200 m + 800 m + 500 m = 1500 m The distance travelled in the first 40 seconds is 1500 m. c) The acceleration when t = 25: Acceleration is the gradient (slope) of the velocity-time graph. Step 1: Identify the segment of the graph at t=25 s. At t=25 s, the graph is a horizontal line segment between t=10 s and t=30 s. Step 2: Calculate the gradient of this segment. In this segment, the velocity is constant at 40 m/s. A horizontal line has a gradient of zero. a = ( v)/( t) = 40 m/s - 40 m/s30 s - 10 s = 0 m/s20 s = 0 m/s^2 The acceleration when t=25 s is 0 m/s^2. 3 done, 2 left today. You're making progress.