As an airplane is taking off at an airport its position is closely monitored by radar. The following three positions are measured with their corresponding times: x1 = 241.22 m at t1 = 3.70 s, x2 = 297.60 m at t2 = 4.20 s, x3 = 360.23 m at t3 = 4.70 s. What is the acceleration of the airplane at t2 = 4.20 s? (Assume that the acceleration of the airplane is constant.)

Answer: acceleration a = 25m/s^2

Explanation:

Given that:

The plane travels with constant acceleration

x1 = 241.22 m at t1 = 3.70 s

x2 = 297.60 m at t2 = 4.20 s

x3 = 360.23 m at t3 = 4.70 s.

We need to calculate the velocity in the two time intervals.

Interval 1:

Average Velocity v1 = ∆x/∆t = (x2 - x1) / (t2-t1)

v1 = (297.60-241.22) / (4.20-3.70) = 112.76m/s

Interval 2:

Average Velocity v2 = ∆x/∆t = (x3-x2) / (t3-t2)

v2 = (360.23-297.60) / (4.70-4.20)

v2 = 125.26m/s

Acceleration:

Acceleration a = ∆v/∆t

∆v = v2-v1 = 125.26m/s-112.76m/s = 12.5m/s

∆t = change in average time of the two intervals = (t3-t1) / 2 = (4.70-3.70) / 2 = 0.5s

a = 12.5/0.5 = 25m/s^2