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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=lv22αv22β (4)
Using equation (1), (2) and (3) we get
t=vα+vβ+l2v (5)
Using (4) and (5), we get
t=vα+vβ+lvv2αv2β
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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