Question

if a particle travelling with the initial velocity of 8 metre per second gets the acceleration of 6 metre per second square and other object travelling with an initial velocity of 20 metre per second the acceleration of 4 metre per second how many seconds later their their velocities will be same​

Answer 1

Let after t seconds their Velocities be equal.

Thus:-

From 1st Equation of Motion we have :-

$a=\dfrac{V-U}{T}$

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CASE 1:-

Initial velocity = 8 m/sec

Acceleration = 6m/sec^2

so,

$6=\dfrac{V-8}{t}\\ \\ \\V-8=6t\\ \\ \\V=6t+8$

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Initial velocity =20m/sec

Acceleration =4m/sec^2

so,

$4=\dfrac{V-20}{t}\\ \\ \\V-20=4t\\ \\ \\V=4t+20$

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Equating the two :-

$6t+8=4t+20\\ \\ \\6t-4t=20-8\\ \\ \\2t=12\\ \\ \\t=\dfrac{12}{2}=6seconds$

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Answer 2

$\mathfrak{ \Large \underline{ \underline{Answer : }}}$

Let t sec be after their velocity be equal.

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Case - 1 :

Initial Velocity = 8 m/s

Acceleration = 6 m/s

So,

$\sf a = \frac{V - U}{t} \\ \to \sf 6 = \frac{V - 8}{t} \\ \to \sf 6t = V - 8 \\ \to \sf V = 6t + 8$

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Case - 2 :

Initial Velocity = 20 m/s

Acceleration = 4 m/s

So,

$\sf a = \frac{V - U}{t} \\ \to \sf 4 = \frac{V - 20}{t} \\ \to \sf 4t = V - 20 \\ \to \sf V = 4t + 20$

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Equating the Case 1 and Case 2 :-

$\sf 6t + 8 = 4t + 20 \\ \to \sf 6t - 4t = 20 - 8 \\ \to \sf 2t = 12 \\ \to \sf t = \frac{12}{2} \\ \to \sf t = 6 \: sec$

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