Identify the final time value on the x-axis.

Here are the solutions based on the velocity-time graph: i) The total time taken for the journey. Step 1: Identify the final time value on the x-axis. The graph starts at t=0 s and ends at t=50 s. The total time taken is 50 s. ii) The maximum velocity reached. Step 1: Identify the highest point on the y-axis (velocity axis) that the graph reaches. The highest velocity shown on the graph is 30 m/s. The maximum velocity reached is 30 m/s. iii) The acceleration during the first 20 seconds. Step 1: Recall that acceleration is the gradient of a velocity-time graph. The acceleration a is given by the formula a = ( v)/( t) = (v_f - v_i)/(t_f - t_i). Step 2: Identify the initial and final velocities and times for the first 20 seconds. Initial velocity (v_i) at t_i = 0 s is 0 m/s. Final velocity (v_f) at t_f = 20 s is 30 m/s. Step 3: Substitute the values into the formula and calculate. a = 30 m/s - 0 m/s20 s - 0 s = 30 m/s20 s = 1.5 m/s^2 The acceleration during the first 20 seconds is 1.5 m/s^2. iv) The total distance travelled. Step 1: Recall that the total distance travelled is the area under the velocity-time graph. The shape under the graph is a trapezoid. The area of a trapezoid is given by A = (1)/(2) (a+b)h, where a and b are the lengths of the parallel sides and h is the height. Step 2: Identify the lengths of the parallel sides and the height from the graph. The parallel sides are the time intervals where the velocity is non-zero. Length of the top parallel side (constant velocity section): 40 s - 20 s = 20 s. Length of the bottom parallel side (total time): 50 s - 0 s = 50 s. The height of the trapezoid (maximum velocity) is 30 m/s. Step 3: Substitute the values into the formula and calculate. Distance = (1)/(2) (20 s + 50 s) × 30 m/s Distance = (1)/(2) (70 s) × 30 m/s Distance = 35 s × 30 m/s Distance = 1050 m The total distance travelled is 1050 m. That's 2 down. 3 left today — send the next one.