Q32E - University Physics with Modern Physics

Question

Two piers, A and B, are located on a river; B is 1500 m downstream from A (Fig. E3.32). Two friends must make round trips from pier A to pier B and return. One rows a boat at a constant speed of   relative to the water; the other walks on the shore at a constant speed of 4.00 km/h. The velocity of the river is   in the direction from A to B. How much time does it take each person to make the round trip?

Step-by-Step Solution

Answer

The time taken by the walker is 0.75 h and the time taken by the person completing the trip by rowing is 1.5 h.

Step 1: Identification of given data:

The given data can be listed below.

• The distance of pier B from A is, d = 1500 m
• The speed of the boat is, v = 4km/h 
• The velocity of the river is, v_r = 2.80km/h 

Step 2: Concept/Significance of relative speed:

Relative velocity is defined as the velocity of an object relative to another observer. It is the time rate of change of relative position of one object with respect to another object.

 The relative speed of two objects traveling in the same direction at different speeds is equal to the difference between their speeds.

Step 3: Determination of time it take each person to make the round trip:

The time taken by the person complete the trip by walking is given by,

$t_w=\frac{d}{v}$

Here, d is the distance traveled by the walker whose values is 2d and v is the velocity of the walker.

Substitute all the values in the above equation.

$\begin{array}{rcl}{t}_w& =& \frac{2\times 1500 m}{4 km/h}\left(\frac{1km}{1000m}\right)\\ & =& 0.75h\end{array}$

The time taken by the person complete the trip by boat is given by,

$\begin{array}{rcl}{t}_p& =& \frac{d}{v_{down}}+\frac{d}{v_{up}}\\ & =& d\left(\frac{1}{v_b+v_r}+\frac{1}{v_b-v_r}\right)\\ & =& \frac{2v_{b}d}{v_b^2-v_r^2}\end{array}$

Here, d is the distance travelled by the walker whose values is 2d and v_b is velocity the of the boat.

Substitute all the values in the above.

$\begin{array}{rcl}{t}_p& =& \frac{2\left(4km/h\right)\left(1.5km\right)}{{\left(4km/h\right)}^2-{\left(2.80km/h\right)}^2}\\ & =& \frac{\left(12k{m}^2/h\right)}{\left(16k{m}^2/h^2\right)-\left(7.84k{m}^2/h^2\right)}\\ & =& 1.5h\end{array}$

Thus, the time taken by the walker is 0.75 h and the time taken by the person completing the trip by rowing is 1.5 h.