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Question

Instantaneous velocity of an object varies with time as v=αβt2. Find its position, x as a function of time, t. Also find the object`s maximum positive displacement, xmax from the origin.

A
x=αtβt33 and xmax=23α3/2β3/2
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B
x=αtβt33 and xmax=2αβ1/2
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C
x=2αtβ and xmax=2αβ
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D
x=2α2tβt3 and xmax=2α1/2β3/2
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Solution

The correct option is A x=αtβt33 and xmax=23α3/2β3/2
Given,
velocity of the object, v=αβt2

From the velocity and displacement relation, we get

v=dxdt=αβt2

dx=(αβt2)dt

Integrating within proper limits, we have

xodx=to(αβt2)dt

x=αt13βt3...(1)

To calculate the maximum value of x i.e., xmax,by differentiating the above equation w.r.t t and equating it to zero, we get

dxdt=αβt2=0

Now, to check whether x is maximum or not at t=αβ, we have to differente dxdt again and substitute the value of t.

d2xdt2=2tβ

[d2xdt2]t= αβ=2αβ

Since, the value of d2xdt2 is negative at t=αβ, so the x will have maximum value at this instant.

Thus, the maximum value of x,

xmax=ααββ3(αβ)3

xmax=α3/2β1/2α3/23.β1/2

xmax=23α3/2β1/2

Hence, option (a) is the correct answer.


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