Question

A cyclist rides 8.9 km east for 18.9 minutes, then he turns and heads west for 4.2 km in 5.9 minutes. Finally, he rides east for 13.4 km, which takes 35.7 minutes. Take east to be the positive direction. What is the displacement (in km)? What is the average velocity (in km/hr)?

Answer 1

Given,

1st displacement of the cyclist, d₁=8.9 km

Time interval in which cyclist covers the 1st displacement, t₁=18.9 min=0.315 hr

2nd displacement, d₂=-4.2 km

Time interval in the cyclist covers the 2nd displacement, t₂=5.9 min=0.098 hr

3rd displacement, d₃=13.4 km

Time interval in which he covers the 3rd displacement, t₃=35.7 min=0.595 hr

We here considered east as the positive direction and the west as the negative direction as mentioned in the question.

To calculate the displacement, we need to find the dum of all the displacement.

Therefore, the displacement is,

`\begin{gathered} D=d_1+d_2+d_3 \\ =8.9+(-4.2)+13.4 \\ =18.1\text{ km} \end{gathered}`

Therefore the total displacement of the cyclist is 18.1 km

The average velocity is given by the displacement dived by the total time duration.

i.e.,

`v=(D)/(t_1+t_2+t_3)`

On substituting the known values in the above equation.

`v=(18.1)/(0.315+0.098+0.595)=17.96\text{ km/hr}`

Therefore the average velocity is 17.96 km.