Question

The Relation between t and distance x is given by
t=ax2 +bx where a and b are constant.
Expreess instaneous acceleration in terms of instaneous velocity. ​

Answer 1

Given, t = ax² + bx

Now, differentiate t with respect to x.

dt/dx = d(ax² + bx)/dx

dt/dx = 2ax + b

Now, Velocity = Displacement/time

v = dx/dt

1/v = dt/dx

From above

1/v = 2ax + b

Now, differentiate v with respect to t (time)

(-1/v²)(dv/dt) = 2a(dx/dt) + 0

(-1/v²)(dv/dt) = 2a(dx/dt)

dv/dt = A and dx/dt = v

So,

(-1/v²)(A) = 2av

-1/v² = (2av)/A

-1/v² = (2av)/A

-1 = (2av³)/A

-A = 2av³

A = -2av³

(Negative sign shows retardation)

Answer 2

$</p><p>\tt{\huge{\pink{Hello!!! }}}$

$</p><p>\tt{\red{Given \: - }}$

t = ax² + bx

$\tt{ \purple{ \implies{ \dfrac{dt}{x} = \dfrac{d(a {x}^{2} + bx) }{dx} }}}$

$\tt{ \blue{ \implies{ \dfrac{1}{v} = \dfrac{dt}{dx} = 2ax \: + \: b \: }}}$

Solving :

$\tt{\red{\implies{A = -2a{v}^3 }}}$

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