Two cars start off a race with velocity `2 ms^(-1)` and `4ms^(-2)` respectively. What is the length of the path if they reach the final point at the same time ?

Explanation

Let both particles reach at same position in same time t then from `s=ut+(1)/(2)at^(2)`
For 1st particle, `s=4(t)+(1)/(2)(1)t^(2)=4t+(t^(2))/(2)`,
For 2nd particle, `s=2(t)+(1)/(2)t^(2)=2t+t^(2)`
Equations, we get `4t+(t^(2))/(2)=2t+t^(2)rArr t=4s`
Substituting value of t in above equations, we get
`s=4(4)+(1)/(2)(1)(4)^(2)=16+8=24m`