The initial velocity of a particle is $10 m/sec$ and its retardation is $2 {m/sec}^{2}$ . The distance covered in the fifth second of the motion will be

1 m

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9 m

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5 m

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7 m

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Solution

The correct option is A $1 m$
Given,
$u = 10 m/s, a = - 2 {m/s}^{2}, n = 5$
As we know that the distance travelled by the particle in $n^{t h}$ second is given as
$S_{n^{th}} = u + \frac{1}{2} \times a (2 n - 1)$
$= 10 + \frac{1}{2} \times (- 2) (2 \times 5 - 1)$
$= 10 - 9$
$= 1 m$

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The initial velocity of a particle is 10 m/sec and its retardation is 2 m/sec2. The distance covered in the fifth second of the motion will be

The correct option is A 1 mGiven, u=10 m/s,a=−2 m/s2,n=5 As we know that the distance travelled by the particle in nth second is given as Snth=u+12×a(2n−1) =10+12×(−2)(2×5−1) =10−9 =1 m

The initial velocity of a particle is $10 m/sec$ and its retardation is $2 {m/sec}^{2}$ . The distance covered in the fifth second of the motion will be

The correct option is A $1 m$
Given,
$u = 10 m/s, a = - 2 {m/s}^{2}, n = 5$
As we know that the distance travelled by the particle in $n^{t h}$ second is given as
$S_{n^{th}} = u + \frac{1}{2} \times a (2 n - 1)$
$= 10 + \frac{1}{2} \times (- 2) (2 \times 5 - 1)$
$= 10 - 9$
$= 1 m$