A body, initially at rest, starts moving with a constant acceleration 2 $m s^{- 2}$ . Calculate:(i) the velocity acquired and (ii) the distance travelled in 5 s.

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Consider a particle initially moving with a velocity of 5 m/s starts decelerating at a constant rate of 2 m/s2. Find the distance travelled in the 2nd second.

The correct option is C 2 mGiven, u=5 m/s,a=−2 m/s2 As we know that the distance travelled by the particle in nth second is given as Sn=u+12a(2n−1) ⇒ S2nd=5+12×(−2)[2(2)−1]=2 m

Consider a particle initially moving with a velocity of $5 m/s$ starts decelerating at a constant rate of $2 m / s^{2}$ . Find the distance travelled in the $2^{n d}$ second.

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