Question

Suppose a plane accelerates from rest for 30 S, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 M down the runway. A smaller plane with the same acceleration has a takeoff speed of 40 M/S. Starting from rest, after what distance will this smaller plane reach its takeoff speed?

A 300 m
B 600 M
C 900 M
D 1200 M

Answer 1

Using the kinematic equation for a uniformly accelerated motion, we deduce from the given parameters that the smaller plane with the same acceleration will reach its takeoff speed of 40 m/s after traveling a distance of 300 meters.

Step-by-step explanation:

The student has posed a question involving the kinematics of a plane accelerating on a runway. Since the acceleration is the same for both planes, we can use the kinematic equations to solve the problem. Specifically, we use the equation v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, s is the distance, and a is the acceleration.

For the first plane, we know that v = 80 m/s, u = 0 (since it starts from rest), and s = 1200 m. We can calculate the acceleration (a) and then apply it to the second plane.

Using the equation, we find that 802 = 2a * 1200, which gives us the acceleration a. We then apply this value to the second plane's scenario where v = 40 m/s. Solving 402 = 2a * s for s, we find the required distance for the smaller plane to reach its takeoff speed.

We find that the distance s is 300 m, so the correct answer is A) 300 m.