Question

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 28.3 m/s. It then flies a further distance of 48900 m, and afterwards, its velocity is 45.5 m/s. Find the airplane's acceleration (m/s^2) and calculate how much time elapses (s) while the airplane covers those 40500 m.

Answer 1

Answer:

a==0.012979m/s^2

t=662.6s

Explanation:

.

Using the Newton equation below

Vf^2 = Vi^2 + 2as

Where Vf= final velocity

Vi= initial velocity

a= acceleration

Vf=45.5 m/s

Vi=28.3 m/s

Xf=48900 m,

Xi=0

a= (Vf^2 - Vi^2)/2s

If we substitute the values,

a=(45.5^2 -28.3^2)/(2×48900)

=(2070.25-800.88)/(2×48900)

=

a= 1269.36/97800

=0.012979m/s^2

Hence, the airplane's acceleration (m/s^2) is =0.012979m/s^2

Using the Newton equation we can calculate the time

a=(vf - vi) /t

Making "t" subject of the formula, we have

t=(vf - vi) /a

a=0.012979m/s^2

If we substitute the values, where

, we have

t= (45.5-28.3)/(2×=0.012979)

=17.2/0.025958

=662.6s

Hence, time elapses (s) while the airplane covers those 40500 m. Is =662.6s