The speed of a train increases at a constant rate `alpha` from zero to v and then remains constant for an interval and finally decreases to zero at a constant rate `beta`. The total distance travelled by the train is l. The time taken to complete the journey is t. Then,

Options

A. `t= (l(alpha + beta))/(alpha beta)`
B. `t= l/v +v/2 (1/alpha + 1/beta)`
C. t is minimum when `v= sqrt((2 l alpha beta)/(alpha - beta))`
D. t is minimum when `v= sqrt((2 l alpha beta)/(alpha + beta))`

Answer

B, D

Explanation

`v=alphat_1 rArr t_1=v/alpha`
`v=betat_2 rArr t_2=v/beta`

`:. t_0=t-t_1=(t-v/alpha-v/beta)`
Now, `l=1/2alphat_1^2+vt_0+1/2betat_2^2`
`=1/2(alpha)(v/a)^2+v(t-v/alpha-v/beta)+1/2(beta)(v/beta)^2`
`=vt-v^2/(2alpha)-v^2/(2beta)`
`:. t=l/v+v/2(l/alpha+1/beta)`
For t to be minimum its first derivation with respect
to velocity be zero or,
`0=-l/v^2+(alpha+beta)/(2alpha beta)`
`:. v=sqrt((2l alphabeta)/(alpha+beta))`