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1st Equation of Motion

A train accel...

Question

A train accelerates from rest at a constant rate $a_{1}$ for distance $S_{1}$ and time $t_{1}$ . After that it decelerates to rest at a constant rate $a_{2}$ for distance $S_{2}$ at time $t_{2}$ . Then the correct relation among the following is

\frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{1}}{t_{2}}

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\frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{1}}{t_{2}}

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\frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}}

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\frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{2}}{t_{1}}

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Solution

The correct option is A $\frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{1}}{t_{2}}$
Given that the particle was at rest at $t = 0$
$s_{1} = u_{1} t + \frac{1}{2} a_{1} t_{1}^{2} . . . . (1)$
$\Rightarrow v_{1} = u_{1} + a_{1} t_{1} . . . . (2)$
$v_{1} \to$ final speed after $t_{1}$
Now, final speed after this is $v_{2}$
And $v_{1} = v_{2}$
$\Rightarrow v_{2} = u_{2} - a_{2} t_{2}$
$\Rightarrow v_{1} = a_{2} t_{2} . . . . (3)$
$\Rightarrow v_{2}^{2} = u_{2}^{2} - 2 a_{2} s_{2}$
$\Rightarrow v_{1}^{2} = 2 a_{2} s_{2}$
From (2), $a_{1}^{2} t_{1}^{2} = 2 a_{2} s_{2}$

$\Rightarrow s_{2} = \frac{a_{1}^{2} t_{1}^{2}}{2 a_{2}}$

$\Rightarrow s_{2} = \frac{a_{1} (2 s_{a})}{2 \frac{a}{2} a_{2}}$

$\Rightarrow \frac{e q^{2} (2)}{e q^{n} (3)} = \frac{a_{1} t_{1}}{a_{2} t_{2}} = 1$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}} = \frac{s_{2}}{s_{1}}$

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A train accelerates from rest at a constant rate a1 for distance S1 and time t1. After that it decelerates to rest at a constant rate a2 for distance S2 at time t2. Then the correct relation among the following is

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