A particle starts its motion from rest under the action of a constant force. If the distance covered in the first 10 seconds is S1 and that covered in the first 20 seconds is S2 , then:

Question

A particle starts its motion from rest under the action of a constant force. If the distance covered in the first 10 seconds is S1 and that covered in the first 20 seconds is S2 , then: (A) S2 = 3 S1 (B) S2 = 4 S1 (C) S2 = S1 (D) S2 = 2 S1

Tardigrade Admin · Accepted Answer

Given: u = 0 , t1 = 10 s , t2 = 20 s Using the relation, S = u t + (1/2) a t2 Acceleration being the same in two cases. S1 = (1/2) a × t12 , S2 = (1/2) a × t22 ∴ (S1/S2) = (. (t1/t2) .)2 = (. (10/20) .)2 = (1/4) S2 = 4 S1

Q. A particle starts its motion from rest under the action of a constant force. If the distance covered in the first $10$ seconds is $S_{1}$ and that covered in the first $20$ seconds is $S_{2}$ , then:

$S_{2} = 3 S_{1}$

$S_{2} = 4 S_{1}$

$S_{2} = S_{1}$

$S_{2} = 2 S_{1}$

Q. A particle starts its motion from rest under the action of a constant force. If the distance covered in the first 10 seconds is S1​ and that covered in the first 20 seconds is S2​ , then:

S2​=3S1​

S2​=4S1​

S2​=S1​

S2​=2S1​

Given: u=0,t1​=10s,t2​=20s Using the relation, S=ut+21​at2 Acceleration being the same in two cases. S1​=21​a×t12​,S2​=21​a×t22​ ∴S2​S1​​=(t2​t1​​)2=(2010​)2=41​ S2​=4S1​

Q. A particle starts its motion from rest under the action of a constant force. If the distance covered in the first $10$ seconds is $S_{1}$ and that covered in the first $20$ seconds is $S_{2}$ , then:

$S_{2} = 3 S_{1}$

$S_{2} = 4 S_{1}$

$S_{2} = S_{1}$

$S_{2} = 2 S_{1}$