Question

An airplane, starting from rest, moves down the runway at constant acceleration for 18 s and then takes off at a speed of 60 m/s. What is the average acceleration of the plane?

Answer 1

Answer:

a =3.33 m/s²

Explanation:

given,

initial speed of Plane, u = 0 m/s

final speed of plane, v = 60 m/s

time of the acceleration, t = 18 s

average acceleration of the plane, a = ?

average acceleration is equal to change in velocity per unit time.

$a = \dfrac{v - u}{t}$

$a = \dfrac{60 - 0}{18}$

$a = \dfrac{60}{18}$

     a =3.33 m/s²

Hence, average acceleration of the plane is equal to a =3.33 m/s²

Answer 2

Answer:

Explanation:

Given

Airplane runs for a time period of $t=18\ s$

Airplane takes off with a velocity of $v=60\ m/s$

It is given that airplane starts from rest i.e. initial velocity $u=0$

acceleration is the rate of change of velocity thus it is given by

$a=\frac{v-u}{t}$

$a=\frac{60-0}{18}$

$a=3.33\ m/s^2$

Average acceleration for 18 s is $3.33 m/s^2$