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Q. 54

Expert-verified
Found in: Page 62

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acceleration, and 5.0 s later the back of the first car passes you. How long does it take after the train starts moving until the back of the seventh car passes you? All cars are the same length.

It will take $13.23s$ to pass the back of the seventh car.

See the step by step solution

## Step 1: Given information

Each car takes time $t=5s$ to pass.

Consider each car has a length of localid="1649230189524" $\mathcal{l}\left(m\right).$

Initial speed of the car, $u=0$

## Step 2: Finding the length of the 7 cars

The distance travelled by a car $l$ can be determined by following the equation $s=ut+\frac{1}{2}a{t}^{2}$as,

localid="1649230197418" $\mathcal{l}=0+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\mathcal{l}=\frac{1}{2}a×{5}^{2}\phantom{\rule{0ex}{0ex}}\mathcal{l}=12.5a$

Therefore, acceleration of each car is, $a=\frac{l}{12.5}m/{s}^{2}.$

Now, length of total seven cars can be determined as,

localid="1649230201861" $S=7\mathcal{l}=0+\frac{1}{2}at{\text{'}}^{2}$

## Step 3: Determination of time

The time taken by the seventh car to pass a person is can be obtained from the equation

$7\mathcal{l}=\frac{1}{2}t{\text{'}}^{2}×\frac{\mathcal{l}}{12.5}\phantom{\rule{0ex}{0ex}}t{\text{'}}^{2}=12.5×2×7\phantom{\rule{0ex}{0ex}}t\text{'}=\sqrt{12.5×2×7}\phantom{\rule{0ex}{0ex}}t\text{'}=13.23s$