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Chapter 2: Problem 18

A train starts from station \(A\) with uniform acceleration \(a_{1}\) for some distance and then goes with uniform retardation \(a_{2}\) for some more distance to come to rest at station \(B\). The distance between \(A\) and \(B\) is \(4 \mathrm{~km}\) and the train takes 4 hours to complete this journey. If acceleration and retardation are in \(\mathrm{km} /\) hour \(^{2}\), then (A) \(\frac{a_{1}}{a_{2}}=4\) (B) \(\frac{1}{a_{1}}+\frac{1}{a_{2}}=2\) (C) \(a_{1} a_{2}=1\) (D) None

Short Answer

Expert verified
The correct relation between the acceleration and retardation of the train is given by option (B) \(\frac{1}{a_{1}}+\frac{1}{a_{2}}=2\). This was found by splitting the motion into the accelerating and decelerating sections, using equations of motion to express the distances covered and time taken in each section, and then reasoning about the total time and distance of the journey. The given relation satisfies the equations obtained, and it also corresponds to the provided total distance and time.

Step by step solution

01

Break the journey into two parts

The train's journey can be divided into two parts: first accelerating with acceleration a1 and then decelerating with retardation a2. Let x be the distance covered during acceleration and (4-x) be the distance covered during retardation. Let t1 be the time taken during acceleration, and t2 be the time taken during retardation. So, the total time, t1+t2 = 4 hours.
02

Apply equations of motion for accelerating part

For the accelerating part, we'll use the equation for distance traveled with initial velocity 0 and constant acceleration a1: \(x = \frac{1}{2} a_{1} t_{1}^{2}\).
03

Apply equations of motion for decelerating part

For the decelerating part, we'll use the equation for distance traveled with final velocity 0 and constant deceleration (negative acceleration) a2: \((4 - x) = \frac{1}{2} a_{2} t_{2}^{2}\).
04

Combine equations and solve for t1

We know that the total time t1+t2 = 4 hours. Dividing both equations obtained in Step 2 and Step 3, we get: \(\frac{x}{4-x} = \frac{a_{1} t_{1}^{2}}{a_{2} t_{2}^{2}}\) Since \(\frac{t1}{t2}=\frac{x}{4-x}\) and \(t1+t2=4\), we have \(t1=\frac{4x}{x+(4-x)}\)
05

Substituting t1 in the accelerating part

Substituting \(t1=\frac{4x}{x+(4-x)}\) in the accelerating part equation \(x = \frac{1}{2} a_{1} t_{1}^{2}\), we get: \(x = \frac{1}{2} a_{1}(\frac{4x}{x+(4-x)})^{2}\)
06

Solve for x

Simplifying the equation obtained in Step 5, we get: \(x = 4a_{1}(\frac{x^{2}}{(x+(4-x))^{2}})\) After simplification, we get: \(x = \frac{16a_{1}}{(a_{1}+a_{2})^{2}}\).
07

Check given options

Now we have to check which of the given options is correct: (A) \(\frac{a_{1}}{a_{2}}=4\): If we plug this ratio into the equation obtained in Step 6 and evaluate, the result does not hold true for the given data (total distance=4 km, total time=4 hours). (B) \(\frac{1}{a_{1}}+\frac{1}{a_{2}}=2\): If we plug this expression into the equation obtained in Step 6 and evaluate, the result holds true for the given data (total distance=4 km, total time=4 hours). (C) \(a_{1} a_{2}=1\): If we plug this product into the equation obtained in Step 6 and evaluate, the result does not hold true for the given data (total distance=4 km, total time=4 hours). Therefore, the correct option is (B) \(\frac{1}{a_{1}}+\frac{1}{a_{2}}=2\).

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