A train accelerates from rest at a constant rate #alpha# for distance X and time T(a).after that it retards at a constant rate #beta# for distance Y and time T(b).Which of the following relations is correct?

Question

#(1) X//Y=alpha//beta=T(a)//T(b)#
#(2X//Y=beta//alpha=T(a)//T(b)#
#(3)X//Y=alpha//beta=T(b)//T(a)#
#(4)X//Y=beta//alpha=T(B)//T(a)#

12863 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-26T14:27:17

#2#

Simply applying the law of kinematics for constant acceleration and retardation,to find the relationship.

For the #1# st half of accelerated motion,we can write,

#v^2 = 2 alpha X# ....1 (#v# is the velocity gained after time #t_a#)

Now,at the end of first #T_a s#, if it gains velocity #v#,then,

#v=0+alpha T_a#

So,putting the value of #v# in equation 1 we get, #alpha ^2 (T_a)^2 = 2 alpha X#

or, #X = alpha (T_a)^2/2#

So,for the rest half of the journey,we can say,

#0^2 = (alpha T_a)^2 - 2 beta Y# or, #Y=alpha ^2 (T_a)^2 /(2 beta)#

And, #0=alphaT_a - beta T_b#

So, #alpha/beta = T_b/T_a#

And, #X/Y = (alpha (T_a)^2)/2 /((alpha)^2 (T_a)^2)/(2 beta)=beta/alpha#

So,we can conclude, #X/Y = beta/alpha = T_a/T_b#