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Chapter 2: Problem 185

A passenger reaches the platform and finds that the second last boggy of the train is passing him. The second last boggy takes \(3 \mathrm{~s}\) to pass the passenger, and the last boggy takes \(2 \mathrm{~s}\) to pass him. If passenger is late by \(t\) second from the departure of the train, then the find the value of \(2 \mathrm{t}\). (Assume that the train accelerates at constant rate and all the boggies are of equal length.)

Short Answer

Expert verified
The value of \(2t\) is 2 seconds.

Step by step solution

01

Identify the variables and write the equations

We have the following variables: - Acceleration of the train: \(a\) - Time taken for the second last boggy to pass the passenger: \(t_1 = 3 \mathrm{s}\) - Time taken for the last boggy to pass the passenger: \(t_2 = 2 \mathrm{s}\) - Time by which the passenger is late: \(t\) - Length of each boggy: \(L\) - Distance covered by the train during the period the passenger was late: \(d\) Since the train accelerates at a constant rate, we can use the equations of motion: 1. For the second last boggy: \(d = L + \frac{1}{2}a{t_1}^2\) 2. For the last boggy: \(d = \frac{1}{2}a({t_2 + t})^2\) We will solve these equations to find the value of \(2t\).
02

Solve the equations to get a relationship between t and L

First, we can rewrite the equation for the second last boggy to express L in terms of d and a: \(L = d - \frac{1}{2}a{t_1}^2\) Now, substitute this expression of L into the equation for the last boggy: \(d = \frac{1}{2}a({t_2 + t})^2 - \frac{1}{2}a{t_1}^2\)
03

Solve for t in terms of d, a and the given times

Now, we can solve this equation for t: \(-\frac{1}{2}a{t_1}^2 = -\frac{1}{2}a{t_2}^2 + at_2t + \frac{1}{2}a{t}^2\) Dividing both sides by \(-\frac{1}{2}a\), we get: \({t_1}^2 = {t_2}^2 - 2t_2t - {t}^2\) Substitute the given values of \(t_1\) and \(t_2\): \((3)^2 = (2)^2 - 2(2)t - t^2\) Simplifying the equation, we get: \(t^2 + 4t - 5 = 0\)
04

Solve the quadratic equation for t

Factorizing the quadratic equation, we get: \((t + 5)(t - 1) = 0\) Therefore, the possible values for t are -5 s and 1 s. However, since t represents the time by which the passenger is late, it cannot be negative. Thus, \(t = 1 \mathrm{s}\).
05

Find the value of 2t

Now, we can find the value of \(2t\): \(2t = 2(1) = 2 \mathrm{s}\) Hence, the value of \(2t\) is 2 seconds.

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