Question

The airplane in the image starts from rest and accelerates east at a constant 3.00 m/s2 for 30.0 s before leaving the ground. a. What was the plane's displacement (Δx)? b. How fast was the airplane going when it took off?

Answer 1

Delta x= (1/2) ^ * a^ * t^ ^ 2

where a is the acceleration and t is the time.

a. Given that the airplane accelerates at a constant 3m / (s ^ 2) for 30.0 s, we can plug these values into the equation:

Delta x=(1/2)^ * 3.00 m/s^ ^ 2^ * (30s) ^ 2

Calculating this, we get:

Delta x=(1/2)^ * 3.00 m/s^ ^ 2^ * 900s ^ 2

Delta*x = 1350m

Therefore, the plane's displacement is 1350.0 meters.

b. To find the speed of the airplane when it took off, we can use the equation:

v =a^ * t where v is the velocity (speed) and a is the acceleration.

Given that the acceleration is 3.00 m / (s ^ 2) and the time is 30.0 s, we can calculate:

v= 3m /s^ ^ 2^ * 30.0 s

v = 90m / s

Therefore, the airplane was traveling at a speed of 90.0 m/s when it took off. I hope I helped you