Chapter 3.2, Problem 35SSC

a.

To determine

The displacement of the plane.

Answer to Problem 35SSC

The displacement of the plane is $\Delta x=1350\text{ m}$ .

Explanation of Solution

Given:

Initial velocity $u=0\text{ m/s}$

Time $t=30\text{ s}$

Acceleration $a{\text{=3 m/s}}^{-2}$

Formula Used:

First equation of motion:

  $v=u+at$

Third equation of motion:

  $v^2=u^2+2as$

v is the final velocity

u is the initial velocity

t is the time

s is the distance

a is the acceleration.

Calculation:

The final velocity is

  $v=u+at$

  $v=0+\left({\text{3 m/s}}^{-2}\right)\left(30\text{ s}\right)$

  $v=90\text{ m/s}$

The displacement is

  $v^2=u^2+2as$

  ${\left(90\text{ m/s}\right)}^2=0^2+2\left({\text{3 m/s}}^{-2}\right)(s)$

  $\Delta x=s=1350\text{ m}$

Conclusion:

The displacement of the plane is $\Delta x=1350\text{ m}$ .

b.

To determine

The takeoff velocity of the plane.

Answer to Problem 35SSC

The takeoff velocity of the plane $v=90\text{ m/s}$ .

Explanation of Solution

Given:

Initial velocity $u=0\text{ m/s}$

Time $t=30\text{ s}$

Acceleration, $a{\text{=3 m/s}}^{-2}$

Formula Used:

First equation of motion:

  $v=u+at$

v is the final velocity

u is the initial velocity

t is the time

a is the acceleration.

Calculation:

The final velocity is

  $v=u+at$

  $v=0+\left({\text{3 m/s}}^{-2}\right)\left(30\text{ s}\right)$

  $v=90\text{ m/s}$

Conclusion:

The takeoff velocity of the plane $v=90\text{ m/s}$ .