Question

At the beginning of a 3.0 -h plane trip, you are traveling due north at $192 \mathrm{km} / \mathrm{h}\(. At the end, you are traveling \)240 \mathrm{km} / \mathrm{h}\( in the northwest direction \)left(45^{\circ}$ west of north). \right. (a) Draw your initial and final velocity vectors. (b) Find the change in your velocity. (c) What is your average acceleration during the trip?

Short answer

Answer: The average acceleration during the plane trip is approximately 53.7 km/h².

Step-by-step solution

(a) Draw initial and final velocity vectors

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).

(b) Find the change in velocity

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

(c) Find the average acceleration during the trip

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.