Question

If a particle starts it's motion from rest under the action of constant force. If the distance covered in first 10 sec is S1 and that covered in first 20sec is the S2 then

Answer 1

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

Answer 2

You can solve the question simply by doing this :

S=ut + ½at²

[u=0]

S1 = ½ * a * (10)² => 50a

S2= ½ * a * (20)² => 200a

Therefore, S2= 4S1

Hope that helps