In terms of Emily's equal steps, what is the length of the ship?

Question

I have a problem here:

2. Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts 210 equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts 42 steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?

I was thinking of turning the steps into seconds. $$ E - B = 210\\ E + B = 42 $$ That's all I got. Please help me!

Answer 1

Let's call Emily's speed $v_E$ and let $v_B$ be the speed of the boat. The length of the boat is $L$. Then assume that Emily starts at one end of the boat, and both move in the same direction. After time $t_1$ Emily walked $210$ steps. In the same time, the boat moved only $210-L$.$$v_E=\frac{210}{t_1}\\v_B=\frac{210-L}{t_1}$$ Taking the ratio of the two equations will cancel out the time:$$ \frac{v_E}{v_B}=\frac{210}{210-L}$$ Now we Emily is moving a time $t_2$ from the other end, in a direction opposite to the boat. At the beginning, the distance to the original end of the boat is $L$. To get there, Emily moves $42$ steps. Therefore the boat moved a distance equal to $L-42$. Applying the same procedure as above: $$v_E=\frac{42}{t_2}\\v_B=\frac{L-42}{t_2}\\\frac{v_E}{v_B}=\frac{42}{L-42}$$ From the two equations involving the $v_E/v_B$ ratio you get $$\frac{210}{210-L}=\frac{42}{L-42}$$ All you need is to rearrange the above equation, to get $L$.

Answer 2

The following time (t) - space (y) diagram describes the question using straight lines representations ($y=st+y_0$ : "present position equals speed times time + initial position"). In particular the bow and the stern of the boat are described by parallel lines with common slope $v_B$. The trajectory of Emily is made of two line segments from the stern to the bow of the boat (slope $v_E$) and then backwards with opposite slope $-v_E$).

Please note that $B'$ is a fictitious point, symmetric of $B$ with respect to $B_2$.

The ratio of speeds of the boat vs. Emilia can be expressed in two ways (remember that in a time-space diagram, the speeds are the slopes):

$$\dfrac{v_B}{v_E}=\dfrac{CC_1}{AC_1}=\dfrac{BB_1}{B'B_1}$$

$$\dfrac{v_B}{v_E}=\dfrac{210-L}{210}=\dfrac{168}{252}$$

a first degree equation with solution

$$L=70$$

Remark: this type of representation is/was used in railway trafic scheduling: see here.

Answer 3

I know the question has been answered but I think I have a simpler explanation.

Let's first assess the movement of the boat in the first instance. The boat and Emily start at the same point, but the end point of the boat must move a further distance $x$ as the walk continues. As we know that Emily moves a distance 210 steps:

$210=L+x$

In the second scenario the end of the boat moves closer to the start point. Since we know that distance is proportional to time (i.e the ratio $T:42:210$), we know that the distance the end point is moved is $\frac{42x}{210}$. Hence we derive:

$42=L-\frac{42x}{210}=L-\frac{x}{5}$

From there you can solve simultaneously.