Question

A helicopter takes off vertically at a speed of 10 meters per second. It then accelerates at a rate of 2.0 m/s squared for 1.5 seconds. How far does it travel in those 1.5 seconds?

Answer 1

The helicopter covers a distance of 17.25 meters during its vertical takeoff and acceleration over a period of 1.5 seconds.

Step-by-step explanation:

To calculate the distance the helicopter travels in 1.5 seconds during its vertical takeoff and acceleration phase, we need to use the kinematic equations of motion. The initial velocity (vi) is 10 meters per second, the time (t) is 1.5 seconds, and the constant acceleration (a) is 2.0 m/s².

Substituting the given values:

• 
  
• d = (10 m/s)(1.5 s) + (1/2)(2.0 m/s²)(1.5 s)²
• 
  
• d = 15 m + (1/2)(2.0 m/s²)(2.25 s²)
• 
  
• d = 15 m + (1)(2.25 m) = 15 m + 2.25 m
• 
  
• d = 17.25 m

Therefore, the helicopter covers a distance of 17.25 meters in those 1.5 seconds.