Question

The relation between time t and distance x is t = ax2 + bx, where a and b are constants. The acceleration is

Answer 1

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Answer 2

GIVEN :–

• Relation between time t and distance(x) is t = ax² + bx, where a and b are constants.

TO FIND :–

• Acceleration = ?

SOLUTION :–

• We know that –

$\\ \: \longrightarrow \: { \boxed{ \sf Acceleration(a) = \dfrac{ {d}^{2}x }{dt {}^{2} }}}$

• And given function –

$\\ \: \implies \: \sf t = a {x}^{2} + bx$

• Now Differentiate with respect to 'x' –

$\\ \: \implies \: \sf \dfrac{dt}{dx} = 2ax \left(\dfrac{dx}{dx} \right)+ b\left(\dfrac{dx}{dx} \right)$

$\\ \: \implies \: \sf \dfrac{dt}{dx} = (2ax + b)$

$\\ \: \implies \: \sf \dfrac{dx}{dt} = \dfrac{1}{(2ax + b)}$

$\\ \: \implies \: \boxed{ \sf velocity = \dfrac{dx}{dt} = \dfrac{1}{(2ax + b)} }$

• We should write this as –

$\\ \: \implies \sf \dfrac{dx}{dt} = (2ax + b)^{ - 1}$

• Now Differentiate again with respect to 't' –

$\\ \: \implies \sf \dfrac{d {}^{2} x}{dt {}^{2} } = ( - 1)(2ax + b)^{ - 2} \left[2a \dfrac{dx}{dt} \right]$

$\\ \: \implies \sf \dfrac{d {}^{2} x}{dt {}^{2} } = \dfrac{ -2a(v)}{(2ax + b)^{ 2}}$

• Put the value of 'v' –

$\\ \: \implies \sf \dfrac{d {}^{2} x}{dt {}^{2} } = \dfrac{ -2a \left( \dfrac{1}{2ax + b}\right)}{(2ax + b)^{ 2}}$

$\\ \: \implies \: \boxed{ \sf Acceleration = \dfrac{d {}^{2} x}{dt {}^{2} } = \dfrac{ -2a}{(2ax+ b)^{3}}}$

USED FORMULA :–

$\\ \sf \: (1) \: \: \: \dfrac{d( {x}^{n}) }{dx} = n {x}^{n - 1}$

$\\ \sf \: (2) \: \: \: \dfrac{d [k.f(x)]}{dx} =k \dfrac{d [f(x)]}{dx}$