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question 17 of 25 a train rolls past a stationary observer. to him, the…

Question

question 17 of 25
a train rolls past a stationary observer. to him, the train is moving at a speed of 23 m/s west, and a woman on the train is moving at a speed of 22.4 m/s west. how long does it take the woman to move 13 m relative to the train?

a. 21.7 s
b. 23.2 s
c. 24.1 s
d. 25.6 s

Explanation:

Step1: Recall speed - time - distance formula

We know that speed $v=\frac{d}{t}$, where $v$ is speed, $d$ is distance and $t$ is time. We want to find $t$, so we can re - arrange the formula to $t = \frac{d}{v}$.

Step2: Identify values of $d$ and $v$

The distance $d$ the woman moves relative to the train is $d = 13$ m. The speed $v$ of the woman relative to the train is $v=22.4$ m/s.

Step3: Calculate time

Substitute $d = 13$ m and $v = 22.4$ m/s into the formula $t=\frac{d}{v}$. So $t=\frac{13}{22.4}\approx0.58$ s. But there seems to be a mistake above. We should use the correct values. Since we want to find the time for the woman to move 13 m relative to the train, and her speed relative to the train is 22.4 m/s. Using $t=\frac{d}{v}$, where $d = 13$ m and $v = 22.4$ m/s, $t=\frac{13}{22.4}\approx0.58$ s. However, if we assume there is a mis - typing in the problem and we use the correct logic with the given values:
We know $v = 22.4$ m/s and $d=13$ m. Using the formula $t=\frac{d}{v}$, we have $t=\frac{13}{22.4}\approx0.58$ s. If we consider the options are wrong and recalculate correctly:
$t=\frac{d}{v}=\frac{13}{22.4}\approx0.58$ s. But if we assume the woman's speed relative to the ground is 22.4 m/s and train's speed relative to the ground is 23 m/s, the woman's speed relative to the train is $v_{woman - train}=23 - 22.4=0.6$ m/s.
Now using $t=\frac{d}{v}$, with $d = 13$ m and $v = 0.6$ m/s, we get $t=\frac{13}{0.6}\approx21.7$ s.

Answer:

A. 21.7 s