Q72P - University Physics with Modern Physics

Question

When a train’s velocity is 12.0m/s eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined $\mathbf{30}\mathbf{°}\mathbf{0}$ to the vertical on the windows of the train. (a) What is the horizontal component of a drop’s velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right)v_{\frac{R}{T}}=-12.0 \frac{m}{s } westward \\ \left(b\right)v_{\frac{R}{E}}=20.80\frac{m}{s }and v_{\frac{R}{T}}=24.0\frac{m}{s } \end{gathered}$$

Step 1: Introduction

From the formula of relative velocity of any object,

 

If any object P is moving related to A and A is moving related to E, then the velocity of P related to E is,

 

$v_{\frac{P}{R}}=v_{\frac{P}{A}}+v_{\frac{A}{R}}$

                                                                                           

Here, $v_{P/E},v_{P/A},v_{A/E}$ are the, velocity of P related to E , velocity of P related to A , and velocity of A related to E respectively.

Step 2: Given

$$\begin{gathered} v_{\frac{R}{T}}=30.0°(west of vertical) \\ {v}_{{}_{\frac{R}{E}}}=vertical with 0\frac{m}{s}horizontal velocity \\ {v}_{{}_{\frac{T}{E}}}=12.0\frac{m}{s}\left(east\right) \end{gathered}$$

Step 3: (a) Horizontal Velocity of raindrop with respect to the earth

The relative velocities of raindrop relative to earth, raindrop related to train, and the train related to earth are $v_{R/E},v_{R/T},v_{T/E}$ respectively. 

 

The relative velocity formula will be,

 

$$\begin{gathered} v_{{}_{\frac{R}{E}}}=vertical with 0\frac{m}{s}horizontal velocity \\ {v}_{{}_{\frac{T}{E}}}=12.0\frac{m}{s}\left(east\right) \\ {v}_{{}_{\frac{R}{E}}}=-v_{{}_{\frac{T}{R}}} \\ {v}_{{}_{\frac{R}{T}}}=-12.0\frac{m}{s}\left(west\right) \end{gathered}$$     

Step 4: (b) The magnitude of raindrop relative to the earth and train

$$\begin{gathered} v_{R/E}=\frac{v_{T/E}}{\tan 30°} \\ {v}_{R/E}=\frac{12.0m/s)}{tan30°} \\ {v}_{R/E}=20.80m/s \end{gathered}$$

 

And further,

 

$$\begin{gathered} v_{\frac{R}{T}}=\frac{v_{T/E}}{sin30°} \\ {v}_{\frac{R}{T}}=\frac{12.0m/s}{sin30°} \\ {v}_{\frac{R}{T}}=24.0m/s \end{gathered}$$