Question

A cyclist travels 5.0 km [S in 45 minuts, then travels 9.0 km [E] in 90 minutes. What is the cyclist's displacement and velocity?

Answer 1

A cyclist travels 5.0 km [S] in 45 minutes, then travels 9.0 km [E] in 90 minutes. What is the cyclist's displacement and velocity?

Solution

The average velocity of a cyclist is

$$v = \frac {s _ {1} + s _ {2}}{t _ {1} + t _ {2}} = \frac {5 \, \text{km} + 9 \, \text{km}}{45 \, \text{min} + 90 \, \text{min}} = \frac {14 \, \text{km}}{135 \, \text{min}} = \frac {14000 \, \text{m}}{135 \cdot 60 \, \text{s}} = 1.7284 \, \frac{\text{m}}{\text{s}}$$

The cyclist's displacement is the vector sum of vectors $\overrightarrow{s_1}$ and $\overrightarrow{s_2}$. Let's find it using Pythagorean theorem (because directions [S] and [E] are perpendicular).

$$S = \sqrt {s _ {1} ^ {2} + s _ {2} ^ {2}} = \sqrt {5 ^ {2} + 9 ^ {2}} = 10.2956 \, \text{km}$$

We can find the direction of cyclist's displacement

$$\tan a = \frac {s _ {2}}{s _ {1}} = \frac {9}{5}; \, a = \tan^ {- 1} \frac {9}{5} = 61{}^{\circ}.$$

So the direction is 61° from south of east or (90-61)=29° from east of south.

Answer: 10.3 km in the direction 29° from east of south; 1.73 $\frac{m}{s}$