**$130\,m$****$160\,m$****$120\,m$****$140\,m$****$150\,m$**

This question aims to choose the correct option out of the five options above, given a scenario.

Kinematics is the discipline of physics that describes motion relative to time and space while neglecting the reason for that motion. Kinematics equations are a collection of equations that can be utilized to calculate an unknown attribute of a body’s motion if the other attributes are known. Kinematic equations are a collection of formulas that characterize an object’s motion with uniform acceleration. Kinematics equations necessitate an understanding of the rate of change, derivatives, and integrals.

These equations can be used to solve a wide range of three-dimensional motion problems involving the object’s motion with uniform acceleration. When solving a problem, a formula should be utilized that includes the unknown variable in addition to three known variables. One parameter is missing in every equation. This enables us to determine which variables are not provided or asked in the problem before choosing the equation which also lacks that variable.

## Expert Answer

To find the velocity of the propeller, first, work out the circumference of its blade as:

$C=\pi r^2$

$C=\pi (12)^2$

$C=144\pi $

Now, $V=\dfrac{C}{t}$

$V=\dfrac{144\pi}{1.2}\,m/s=120\pi\, m/s$

Now the total distance is $d=32\,m$, $a=9.8\,m/s^2$ and $V_0=0$, therefore:

$d=V_0t+\dfrac{1}{2}at^2$

$32=0+\dfrac{1}{2}(9.8)t^2$

$32=4.9t^2$

$t^2=6.53\,s^2$

$t=2.55\,s$

Let $x$ be the distance from the base of the pylon to the point where the fragment strikes the ground, then:

$x=\dfrac{120\pi}{2.55}$

$x=\dfrac{120\pi}{2.55}=147.8\,m$

## Example 1

A plane accelerates down a runway at $2.12 \,m/s^2$ for $23.7$ seconds before lifting off. Calculate the distance traveled prior to takeoff.

### Solution

Given that:

$a=2.12\,m/s^2$, $t=23.7\,s$ and $v_0=0$.

Using the distance formula:

$d=V_0t+\dfrac{1}{2}at^2$

$d=(0)(23.7)+\dfrac{1}{2}(2.12)(23.7)^2$

$d=0+595.39$

$d=595\,m$

## Example 2

A car begins at rest and uniformly accelerates in $2.5\,s$ for a distance of $221\, m$. Evaluate the car’s acceleration.

### Solution

Given that:

$d=221\, m$, $t=2.5\,s$ and $v_0=0$.

Using the distance formula:

$d=V_0t+\dfrac{1}{2}at^2$

$221=(0)(2.5)+\dfrac{1}{2}a(2.5)^2$

$221=0+3.125a$

$221=3.125a$

$a=70.72\,m/s^2$