A train accelerates from rest at a constant rate a for distance `x_(1)` and time `t_91)`. After that is retards at constant rate `beta` for distance `x_(2)` and time `t_(2)` and comes to the rest. Which of the following relation is correct:-

Options

A. `(x_(1))/(x_(2))=(alpha)/(beta)=(t_(1))/(t_(2))`
B. `(x_(1))/(t_(2))=(beta)/(alpha)=(t_(1))/(t_(2))`
C. `(x_(1))/(x_(2))=(alpha)/(beta)=(t_(2))/(t_(1))`
D. `(x_(1))/(x_(2))=(beta)/(alpha)=(t_(2))/(t_(1))`

Answer

B

Explanation

`tan alpha="slope"=(v_("max"))/(t_(1))rArr alpha=tan^(-1)((v_("max"))/(t_(1)))`
`tan beta="slope"= = (v_("max"))/(t_(2)) rArr beta=tan ^(-1)((V_("max"))/(t_(2)))`
`therefore (beta)/(alpha)=((v_("max"))/(t_(2)))/(v_("max")/(t_(1)))=(t_(1))/(t_(2))`
Area of (v-t) graph= displacement
During acceleration graph `x_(1)=(1)/(2)xxt_(1)v_("max")` ....(1)
During retarding graph `x_(2)=(1)/(2)xxt_(2)v_("max")`....(2)
Now `(x_(1))/(x_(2))=(t_(1))/(t_(2))` so `(x_(1))/(x_(2))=(t_(1))/(t_(2))=(beta)/(alpha)`