Correct Answer - D
` x_A=A_b` :. Pt+qt^2 =rt^2 -st^3`
One solution is ` t= 0`, which means both the cars start from the same point. To find the other solution , divede it by ` 9t), we have ` p + qt =rt -st^2` or ` st^2 + (q-2) t + p=0`
or ` t= 1/(2 s) [ q- r) =- sqrt((q -r)^2 -4sp)]`
` =1/(2 xx 0.20) [1 . 20 - 2 . 89 )`
` +- sqrt ((1.20 - 2 .80)^2 - 4 xx 0.24 xx 2.60)]`
`= 4. 00 +- 73 = 2.27 s ` and ` 5. 73 s`.
Hence ` x_A= x_B` for ` t= 0, t= 2.27 s` and ` t= 5. 73 s`.