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Question

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

A
t=l(α+β)αβ
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B
t=lv+v2(1α+1β)
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C
t is minimum when v =2tαβ(αβ)
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D
t is minimum when v =2lαβ(α+β)
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Solution

## The correct options are B t=lv+v2(1α+1β) D t is minimum when v =√2lαβ(α+β)At t=0, u=0 From the figure t=t1+t2+t3 (1) By using first equation of motion, we get t1=va and t3=vβ (2) Let l2 be the distance covered with constant velocity ⇒t2=l2v (3) By using third equation of motion: also , l1=v22α and l3=v22β According to question, l=l1+l2+l3 ∴l2=l−v22α−v22β (4) Using equation (1), (2) and (3) we get t=vα+vβ+l2v (5) Using (4) and (5), we get t=vα+vβ+lv−v2α−v2β ∴t=lv+v2(1α+1β) For t to be minimum dtdv=0 −lv+12(β+ααβ)=0 ⇒v=√2lαβ(α+β)

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