Since it is an upward drawn parabola ...the value of f for those values between the roots (-2 and 2 minimum) are negative

so f(1)<0 =>a+b+c<0

f(-1)<0 =>a-b+c<0

Thus a+|b|+c<0

Assume it is displacd by x straight wrt to one spring...then correspondingly the extensions in the other two springs will be xcos(60) each

The force eqn (which is not much different,would be)

 

F = kx+(k²x²cos²(60)+k²x²cos²(60)+2k²x²cos²(60)cos(120))

(Using the formula of vectors .a+b(vectors)(magnitude) = (a²+b²+2abcos(alpha)

where alpha is the angle between the vectors a and b

 

Thus F = kx+kx/2 =3kx/2

T = 2(m/k_eff) where k_eff in this case is 3k/2

Thus T = 2(2m/3k)

I suppose everybody is very serious about VITEEE because the same question has come thrice in 3 days

 

Here is the link

 

The important point is that it is a rod and not a string ..a string can slack but a rod cannot

Thus the least condition is that the topmost pt ..the velocity can be 0 .

thus mg(2l) = 1/2 m(v²)

which means v = 4gl

Ryu ..i guess u've made a small mistake...

Question says ...each cat has 7 kittens ...

so u actually have 7 cats +7*7 kittens = 56 cats!

I m assuming that the river velocity is perpendicular to the direction in which the boatman wants to move ...to be more clear ..consider the 4 known directions ..north,south,east ,west.

Assume the boatman wants to move from South to North, and let the river flow be from west to east.

 

First part of the question ..suppose he wants to reach exactly opp. to where he started..

He should try to move in the north west dir..so that its horizontal component cancels the rightward velocity of the river...

Let the angle made by his velocity with the horizontal line be a.then

vcos(a) = 5 => 10cos(a) = 5 => a = 60 degrees ...Thus to reach the exactly opp pt. he should row at 60 degrees with the horizontal line ..in this case ..west -east line ..

 

Now if he has to reach the bank in the shortest possible time ...then there is only way to do it..row straight ...

Note that the horizontal velocity does not affect our discussion ..it is only the forward velocity that matters ....and we know the maximum forward velocity is his speed 10kmph ...This way he'll reach fastest ..

 

Now assume he is going at an angle 60 degrees(according to first case)

Then effective forward velocity = 10sin(60) = 103/2 = 53

Thus time taken is distance/speed = 1/53 hrs =6.928 minutes

 

Now assume he is going straight ...this case vertical velocity is 10kmph ...

Thus time taken is 1/10 hrs = 6 minutes