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Standard XII

Physics

Position Vector

A particle is...

Question

A particle is moving with a velocity $35 m / s$ along positive $x$ -axis, it's acceleration is towards negative x-axis with the magnitude $4 m / s^{2}$ then find the distance covered by the particle in the $9 t h$ second.

1 m

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\frac{9}{8} m

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\frac{5}{4} m

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153 m

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Solution

The correct option is A $1 m$
$Given :$ Initial Velocity, u = 35 m/s
Acceleration, a = $- 4 m / s^{2}$

$Solution :$
The distance covered by the particle in nth second is given by the formula,
$s_{n} = u + (\frac{a}{2}) (2 n - 1)$
$⟹ s_{9} = 35 + (\frac{- 4}{2}) (2 \times 9 - 1)$
$⟹ s_{9} = 35 - 2 \times 17$
$⟹ s_{9} = 35 - 34$
$⟹ s_{9} = 1 m$

$Hence A is the correct option$

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A particle is moving with a velocity 35 m/s along positive x-axis, it's acceleration is towards negative x-axis with the magnitude 4 m/s2 then find the distance covered by the particle in the 9th second.

The correct option is A 1 mGiven: Initial Velocity, u = 35 m/s Acceleration, a = −4m/s2Solution:The distance covered by the particle in nth second is given by the formula,sn=u+(a2)(2n−1)⟹s9=35+(−42)(2×9−1)⟹s9=35−2×17⟹s9=35−34⟹s9=1mHence A is the correct option

A particle is moving with a velocity $35 m / s$ along positive $x$ -axis, it's acceleration is towards negative x-axis with the magnitude $4 m / s^{2}$ then find the distance covered by the particle in the $9 t h$ second.