Question

Rain is falling vertically with a speed of 35m/s.A women rides a bicycle with a speed of 12m/s in east to west direction what is the direction in which she should hold her umbrella

Answer 1

Answer:

71.1o relative to the horizontal

Explanation:

For this question, we'll be finding the relative velocity of the rain with respect to the woman.

The relative velocity of the rain relative to the woman (denoted →vR/W) is given by

→vR/W=→vR/E−→vW/E−−−−−−−−−−−−−−−−−−−−−

where

→vR/E is the velocity of the rain with respect to the Earth →vW/E is the velocity of the woman with respect to the Earth

We'll separate this equation into component form:

vRx/Wx=vRx/E−vWx/E vRy/Wy=vRy/E−vWy/E

I'll call the positive x-direction west (the direction the woman is traveling) and the positive y-direction north.

With this in mind, we have:

vRy/E=−35 m/s (velocity of rain relative to Earth) vWx/E=12 m/s (velocity of woman relative to Earth)

(The other components are 0, because the rain moves only in the y-direction, and the woman moves only in the x-direction.)

Plugging the components into the equations:

vRx/Wx=0−12lm/s=−12lm/s−−−−−−−− vRy/Wy=−35lm/s−0=−35lm/s−−−−−−−−

The direction of the relative velocity of the rain from the woman's perspective is given by

tanθ=vRy/WyvRx/Wx

So

θ=arctan(−35lm/s−12lm/s)=71.1o+180o=251o−−−− (the rain is falling in the third quadrant relative to the woman [because both the x- and y-components are negative], so the 180o was added to fix the calculation error)

The direction (relative to the horizontal) she must hold her umbrella is the opposite of this angle (because this represents the angle the rain travels with respect to the woman).

Thus, the direction the woman must hold her umbrella relative to the horizontal is

251o−180o=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ 71.1o ∣∣−−−−−−−−−−

which is what the calculator showed the value for the arctangent was.