A particle starts moving with acceleration 2 m / s 2 . Distance travelled by it in 5th half second is

Question

A particle starts moving with acceleration 2 m / s 2 . Distance travelled by it in 5th half second is (A) 1.25 m (B) 2.25 m (C) 6.25 m (D) 30.25 m

Tardigrade Admin · Accepted Answer

S2.5-S2=? (distance travelled in 5th half second) S2.5=u t+(1/2) a t2 a=2 ms -2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/30242ae332ac246fdf634ab6f3b868b7-.png" /> ⇒ S2.5=(1/2) × 2 ×(2.5)2=6.25 m (∵ u=0) S2=(1/2) × 2 × 4=4 m So, S2.5-S2=2.25 m

Q. A particle starts moving with acceleration $2 m / s^{2} .$ Distance travelled by it in $5^{t h}$ half second is

1.25 m

2.25 m

6.25 m

30.25 m

$S_{2.5} - S_{2} = ?$ (distance travelled in $5^{t h}$ half second)
$S_{2.5} = u t + \frac{1}{2} a t^{2}$
$a = 2 m s^{- 2}$

$\Rightarrow S_{2.5} = \frac{1}{2} \times 2 \times (2.5)^{2} = 6.25 m$
$(∵ u = 0)$
$S_{2} = \frac{1}{2} \times 2 \times 4 = 4 m$
So, $S_{2.5} - S_{2} = 2.25 m$

Q. A particle starts moving with acceleration 2m/s2. Distance travelled by it in 5th half second is