Question

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 degrees above the x axis. What was the magnitude of the acceleration of the spaceship during the 25 seconds?

A) 1.5 m/s^2
B) 7.3 m/s^2
C) 28 m/s^2
D) 48 m/s^2
E) 57 m/s^2

Answer 1

The magnitude of the acceleration of the spaceship over 25 seconds is 7.3 m/s², calculated by first determining the x and y components of the final velocity and using these to find the change in velocity components. The acceleration components are then found and combined using the Pythagorean theorem.

Step-by-step explanation:

A spaceship accelerates from an initial speed of 150 m/s to a speed of 1500 m/s over the course of 25 seconds at an angle of 55 degrees above the x-axis. To find the magnitude of the acceleration, we need to consider the components of the final velocity and use these to find the change in velocity over time.

First, we find the x and y components of the final velocity:

• Vx_final = 1500 m/s × cos(55°)
• Vy_final = 1500 m/s × sin(55°)
• Then, we find the change in each component of the velocity:

• ΔVx = Vx_final - Vx_initial
• ΔVy = Vy_final - 0 (since it starts with no y-component)
  Next, we calculate the components of the acceleration :

• ax = ΔVx / time
• ay = ΔVy / time
  Finally, we find the magnitude of the acceleration by using the Pythagorean theorem:

• a = √(ax² + ay²)

After calculating, the required answer is B) 7.3 m/s².