The relation between time t and distance x is $t = a x^{2} + b x$ where a and b are constants.The acceleration is

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Answer 1

$t = a x^{2} + b x$
Differentiate w.r.t time
$1 = 2 a x \frac{d x}{d t} + b \frac{d x}{d t}$
$\frac{d x}{d t} = \frac{1}{2 a x + b}$
$V = \frac{1}{2 a x + b}$ ...(1)
$a = \frac{d V}{d t} = \frac{- ( 2 a \frac{d x}{d t} )}{( 2 a x + b ) ^{2}}$
$a = \frac{- 2 a v}{( 2 a x + b ) ^{2}}$ ....(2)
Substiting $\frac{1}{2 a x + b} = V$ in (2)
we will get, $a = - 2 a v^{3}$

Answer 2

t = a x^{2} + b x

Differentiate w.r.t time

1 = 2 a x \frac{d x}{d t} + b \frac{d x}{d t}

\frac{d x}{d t} = \frac{1}{2 a x + b}

V = \frac{1}{2 a x + b}

...(1)

a = \frac{d V}{d t} = \frac{- ( 2 a \frac{d x}{d t} )}{( 2 a x + b ) ^{2}}

a = \frac{- 2 a v}{( 2 a x + b ) ^{2}}

....(2)

Substiting

\frac{1}{2 a x + b} = V

in (2)

we will get,

a = - 2 a v^{3}

The relation between time t and distance x is $t = a x^{2} + b x$ where a and b are constants.The acceleration is

$- 2 a v^{3}$

$2 a v^{2}$

$- 2 a v^{2}$

$2 b v^{3}$

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$c_{1}$ and $c_{2}$ are positive constants The acceleration of the particle is

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

−2av3

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t=ax2+bxDifferentiate w.r.t time1=2axdtdx​+bdtdx​dtdx​=2ax+b1​V=2ax+b1​...(1)a=dtdV​=(2ax+b)2−(2adtdx​)​a=(2ax+b)2−2av​....(2)Substiting 2ax+b1​=V in (2)we will get, a=−2av3

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$- 2 a v^{3}$

$2 a v^{2}$

$- 2 a v^{2}$

$2 b v^{3}$

The velocity where $v$ of a particle as a function of its position (x) is expressed as $v = c_{1} - c_{2} x,$
$c_{1}$ and $c_{2}$ are positive constants The acceleration of the particle is

Fill in the blanks :
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The relation between time $t$ and distance $x$ is $t = a x^{2} + b x$ where a and b are constants. The acceleration is

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