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Q.

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

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a

t=l(α+β)αβ

b

t=lv+v21α+1β

c

t is minimum when v=2lαβ(α-β)

d

t is minimum when v=2lαβ(α+β)

answer is B.

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Detailed Solution

ν=αt1⇒t1=ναv=βt2⇒t2=vβ∴ t0=t-t1-t2=t-vα-vβNow, l=12αt12+vt0+12βt22=12(α)vα2+vt-vα-vβ+12(β)vβ2=vt-v22α-v22β∴ t=lv+v21α+1βFor t to be minimum its first derivation with respect to velocity be zero, i.e.0=-1v2+α+β2αβ∴v=2lαα+β

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