A spaceship is observed traveling in the positive x direction with a speed of 150 m/s when it begins accelerating at a constant rate. The spaceship is observed 25 s later traveling with an instantaneous velocity of 1500 m/s at an angle of 55° above the +x axis.
-What was the magnitude of the displacement of the spaceship during the 25 seconds?

Question

Quizplus · Accepted Answer

The correct answer is A.The displacement of the spaceship is the vector difference between its final and initial positions. The initial position is at the origin, and the final position is at (1500 m/s) * (25 s) * cos(55°) = 2.0 × 10^4 m in the x direction and (1500 m/s) * (25 s) * sin(55°) = 1.4 × 10^4 m in the y direction. The magnitude of the displacement is therefore sqrt((2.0 × 10^4 m)^2 + (1.4 × 10^4 m)^2) = 2.4 × 10^4 m.

A Spaceship Is Observed Traveling in the Positive X Direction