Question

t=ax^2+bx find velocity of the particle by method of differentiation​

Answer 1

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t=ax^2+bx find velocity of the particle by method of differentiation

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$t \: = \: {ax}^{2} + \: bx$

$differentiate \: \: w.r.t \: \: time$

$1 = 2 ax\frac{dx}{dt} + b\frac{bx}{dt}$

$\frac{dx}{dt} = \frac{1}{2ax + b}$

$v = \frac{1}{2ax + b} ...(1)$

$a \: = \: \frac{dv}{dt} = \frac{( - 2a \frac{dx}{dt)} }{(2ax + {b)}^{2} }$

$a = \frac{ - 2av}{(2ax + {b)}^{2} }...(2)$

$substiting \frac{1}{2ax + b} = v \: \: in \: \: 2$

$we \: will \: get \: a \: = - {2av}^{3}$