A particle experiences constant acceleration for $$20$$ seconds after starting from rest. If it travels a distance $${s_1}$$ in the first $$10$$ seconds and distance $${s_2}$$ in the next $$10$$ seconds, then

Solution :
Let $$a$$ be the constant acceleration of the particle. Then
$$s = ut + \frac{1}{2}a{t^2}\,\,{\text{or}}\,\,{s_1} = 0 + \frac{1}{2} \times a \times {\left( {10} \right)^2} = 50\,a$$
$$\eqalign{ & {\text{and}}\,{s_2} = \left[ {0 + \frac{1}{2}a{{\left( {20} \right)}^2}} \right] - 50a = 150a \cr & \therefore {s_2} = 3{s_1} \cr} $$