How far does a plane fly in $15$ s while its velocity is increasing from $75 m s^{-1}$ to $145 m s^{-1}$ at a uniform rate of acceleration?

Hint: We have to find how far a plane can fly, it means we need to find distance, how much distance can a plane cover with the change of velocity in the given time. Time is given; initial and final velocities are given so we will find the acceleration of the plane by the first equation of motion. With the help of a third equation of motion, we will find the distance travelled by the plane. That distance represents how far a plane can fly in a given time when its velocity changes from initial to final value.

Complete step by step answer:
 Given: initial velocity, $u = 75 m s^{-1}$
Final velocity, $v = 145 m s^{-1}$
Time taken = $15 s$.
According to the first equation of motion, we will find the acceleration.
$v = u +at$
$\implies 145 = 75 +a(15)$
$\implies a =\dfrac{70}{15} = 4.67 m s^{-2}$
According to the third equation of motion, we will find the distance.
$v^{2} = u^{2} + 2as$
$\implies 145^{2} = 75^{2} + 2(15)s$
$\implies s = 1648.8 m$

Hence, the distance is $1648.8 m$.

Note: All the values should be in S.I. unit. When velocity increases, then rate of velocity change is positive and when velocity is decreasing then rate of velocity change is negative. If the velocity change is at uniform rate, the body is said to be in uniform acceleration.