# Ship A leaves port sailing north at a speed of 20 mph. A half hour later, Ship B leaves the same...

## Question:

Ship A leaves port sailing north at a speed of 20 mph. A half hour later, Ship B leaves the same port sailing east at a speed of 35 mph. Let {eq}t {/eq} (in hours) denote the time ship B has been at sea.

a) Find an expression in terms of {eq}t {/eq} giving the distance {eq}D {/eq} between the two ships.

b) Use the expression obtained in part (a) to find the distance between the two ships 4 hours after A has left port.

## Word problem involving distance and speed:

Here we are given a word problem which involves

(a) obtaining an expression for distance in terms of t, given the speed.

(b) calculating the distance between two moving bodies at a particular time, given their speed, direction and starting point.

## Answer and Explanation:

(a)

Ship A leaves port sailing north at a speed of 20 mph. A half hour later, Ship B leaves the same port sailing east at a speed of 35 mph.

We can take the port to be the origin , then A is travelling along the Y axis and B is travelling along X axis.

{eq}t {/eq} hours denotes the time that B has been at sea.

So A has been at sea for {eq}t + 0.5 {/eq} hours.

The distance travelled by A :

{eq}20 \times ( t+0.5) = 20 t + 10 {/eq} m.

The distance travelled by B:

{eq}35 \times t {/eq} m.

Since A is travelling along the Y axis and B along the X axis, at time {eq}t {/eq} their coordinates are :

{eq}A : ( 0, 20t +10)\\ B: ( 35t,0) {/eq}

Thus using the formula for the distance between two points {eq}(x_1,y_1) \; \text{and}\; (x_2, y_2)\\ \displaystyle{ \sqrt(x_2 - x_1)^2 + (y_2 -y_1)^2 } {/eq}

we have :

{eq}\displaystyle{ \begin{align} D &= \sqrt{ (35t)^2 + (20t + 10)^2}\\ &= \sqrt{ 1625 t^2 + 400t + 100}\\ &= 5 \sqrt{ 65t^2 + 8t +4} \end{align} } {/eq}

(b)

4 hours after A leaves the port means it is 3.5 hours after B leaves the port. So {eq}t = 3.5 {/eq} hours.

Thus

{eq}\displaystyle{ \begin{align} D(3.5) &= 5 \sqrt { 65 (3.5)^2+ 8(3.5) + 4}\\ &= 5 \sqrt { 796.25 + 28 + 4}\\ &= 5\sqrt{828.25}\\ &= 5 \times 28.78\\ &= 143.90\; m \end{align} } {/eq}

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from 6th-8th Grade Math: Practice & Review

Chapter 32 / Lesson 8