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

a. S2=2S1

b. S2=3S1

c. S2=4S1

d. S2=S1

We will be using kinematics equation for uniform acceleration type motion,
s=ut+1/2at2
u= initial velocity = 0m/s

s= distance
assume the acceleration produced due to the said force be A.
for the first 10 seconds,
S1=1/2a(10)2
At the end of the 10th second, The final velocity will be treated as the initial velocity for the S2 part of journey.
Calcuating final velocity during the first 10 seconds,
v=u+at
v=10a
Now this v is the initial velocity for S2 journey,
Hence,
S2=10a+1/2a(10)2
S2=4S1

option c is correct