Two objects start at different positions or times with different accelerations/velocities, and the question asks when or where they meet, have equal velocities, or one catches the other.
Two points $A$ and $B$ move from rest along a straight line with constant acceleration $f$ and $f'$ respectively. If $A$ takes $m$ sec. more than $B$ and describes '$n$' units more than $B$ in acquiring the same speed then (1) $\left(f - f'\right)m^2 = ff'n$ (2) $\left(f + f'\right)m^2 = ff'n$ (3) $\frac{1}{2}\left(f + f'\right)m = ff'n^2$ (4) $\left(f' - f\right)n = \frac{1}{2}ff'm^2$
A ball is projected vertically upward with an initial velocity of $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $t = 0 \mathrm {~s}$. At $t = 2 \mathrm {~s}$, another ball is projected vertically upward with same velocity. At $t =$ $\_\_\_\_$ s, second ball will meet the first ball $\left( \mathrm { g } = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 } \right)$.
Two balls $A$ and $B$ are placed at the top of 180 m tall tower. Ball $A$ is released from the top at $t = 0 \mathrm {~s}$. Ball $B$ is thrown vertically down with an initial velocity $u$ at $t = 2 \mathrm {~s}$. After a certain time, both balls meet 100 m above the ground. Find the value of $u$ in $\mathrm { m } \mathrm { s } ^ { - 1 }$. [use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$] (1) 10 (2) 15 (3) 20 (4) 30