Question

A jet accelerates from rest down a runway at 1.75 m/s2 for a distance of 1500 m before

takeoff.
a. How fast is the plane moving at takeoff?
b. How long does it take the plane to travel down the runway?

Answer 1

Step-by-step explanation:

a. Using the third equation of motion:

v2 = u2 + 2as

from the question;

the jet was initially at rest

hence u = 0

a = 1.75m/s2

s = 1500m

v2 = 02 + 2(1.75)(1500)

v2 = 5250

v = √5250

v = 72.46m/s

hence it moves with a velocity of 72.46m/s.

b. s = ut + 1/2at2

1500 = 0(t) + 1/2(1.75)t2

1500 × 2 = 2× 1/2(1.75)t2

3000 = 1.75t2

1714.29 = t2

41.4 = t

hence the time taken for the plane to down the runway is 41.4s.