Question

A particle starting from rest experiences a constant acceleration for 20
seconds. If it travels a distance y, in the first 10 sec and a distance y2
in the next 10 sec. then obtain a relation between y, and y2.​

Answer 1

As acceleration is constant for 20 sec after starting from rest so

S1=ut +1/2 ×at^2

S1=0×10 +1/2 ×a×100

So,S1=50a

Now,S1+S2=0×20+1/2 ×a×20^2

S1+S2=200a

As, S1=50a

50a +S2=200a

S2=150a

Hence S1/S2=50a/150a

S2=3S1