Question

Objects detected by radar are plotted in a coordinate plane where each unit represents $1mile$. The point $(0,0)$represents the location of a shipyard. A cargo ship is traveling at a constant speed and in a constant direction parallel to the coastline. At $9A.M.$, the radar shows the cargo ship at $(0,15)$. At $10A.M.$, the radar shows the cargo ship at $(16,15)$. How far is the cargo ship from the shipyard at $4P.M.$? Explain.

Answer 1

Step 1: Find the distance covered by the cargo ship till $4P.M.$

We have given that, objects detected by radar are plotted in a coordinate plane where each unit represents $1mile$. The point $(0,0)$represents the location of a shipyard. A cargo ship is traveling at a constant speed and in a constant direction parallel to the coastline. At $9A.M.$, the radar shows the cargo ship at $(0,15)$. At $10A.M.$, the radar shows the cargo ship at $(16,15)$.

So, we can say, In one hour the ship covers $16miles$ as it moves from position $(0,15)$ to $(16,15)$ in one hour. and use the formula of speed to find the speed of the cargo ship

$Speed=\frac{dis\tan ce}{time}=\frac{16miles}{1hour}=16miles/hour$

At $4pm$ it covers $7hours$in total

So, $Dis\tan ce=speed\times time=16\times 7=112miles$

Step 2:Find the distance of the cargo ship from the shipyard at $4P.M.$

The distance of the cargo ship from the shipyard is the hypotenuse of the triangle formed whose legs are $15$ and $112$.

Using Pythagoras theorem

$a^2+b^2=c^2$

Substitute $a=15$ and $b=112$.

$$\begin{gathered} {15}^2+{112}^2=c^2 \\ \Rightarrow 225+12544=10.24 \\ \Rightarrow 12769=c^2\mathbf{} \\ \Rightarrow \sqrt{12769}=c\mathbf{}\mathbf{(}\mathbf{Take}\mathbf{}\mathbf{positive}\mathbf{}\mathbf{square}\mathbf{}\mathbf{root}\mathbf{}\mathbf{of}\mathbf{}\mathbf{each}\mathbf{}\mathbf{side}\mathbf{)} \\ \Rightarrow 113=c \\ \therefore c=113miles. \end{gathered}$$

Hence, the distance of the cargo ship from the shipyard at $4P.M.$ is $113miles$.