91Ó°ÊÓ

To draw the initial and final velocity vectors, we first need to understand the directions: 1. Initial velocity: Due north, with a magnitude of 192 km/h. 2. Final velocity: Northwest direction (45° west of north), with a magnitude of 240 km/h. Now we can draw the vectors on a coordinate plane: 1. Initial velocity (Vi): Pointing upwards (north) with a length of 192 units (representing km/h). 2. Final velocity (Vf): Pointing in an angle of 45° west of north with a length of 240 units (representing km/h).

To find the change in velocity, we need to subtract the initial velocity vector from the final velocity vector. To do this, we can break down the vectors into their x and y components: Initial velocity (Vi) components: Vi_x = 0 (due north) Vi_y = 192 km/h Final velocity (Vf) components: Vf_x = 240 * cos(45°) = 240 * \frac{\sqrt{2}}{2} km/h = 120 \sqrt{2} km/h Vf_y = 240 * sin(45°) = 240 * \frac{\sqrt{2}}{2} km/h = 120 \sqrt{2} km/h Now, we can find the change in velocity (∆V) components: ∆V_x = Vf_x - Vi_x = 120 \sqrt{2} km/h ∆V_y = Vf_y - Vi_y = 120 \sqrt{2} km/h - 192 km/h Now, we find the magnitude of the change in velocity using the Pythagorean theorem: ∆V = \sqrt{(∆V_x)^2 + (∆V_y)^2} = \sqrt{(120 \sqrt{2} km/h)^2 + (120 \sqrt{2} km/h - 192 km/h)^2} ∆V ≈ 161.2 km/h

To find the average acceleration (a_avg), we need to divide the change in velocity (∆V) by the time interval (∆t). The time interval is given as 3.0 h. a_avg = \frac{∆V}{∆t} = \frac{161.2 km/h}{3.0 h} ≈ 53.7 km/h² Thus, the average acceleration during the trip is 53.7 km/h².