Suppose a plane accelerates from rest for 32.3, achieving a takeoff speed of 47.1 m/s after traveling a distance of 607 m down the runway. a smaller plane with the same acceleration has a takeoff speed of 28.2 m/s. starting from rest, after what distance will this smaller plane reach its takeoff speed?

217.28 m/s

Explanation:

u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m

Let a be the acceleration.

Use third equation of motion.

v^2 = u^2 + 2 a s

47.1 x 47.1 = 0 + 2 a x 607

a = 1.83 m/s^2

For small plane

a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.

Use third equation of motion

28.2^2 = 0 + 2 x 1.83 x s

s = 217.28 m/s

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