The relation between time and distance is t = | vedclass

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Question

(a) $\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}$

$a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{

Vedclass · Accepted Answer

$2\alpha {v^3}$

Vedclass · Answer

(a) $\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}$

$a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}$

$a = v\frac{{dv}}{{dx}} = \frac{{ - v.2\alpha }}{{{{(2\alpha x + \beta )}^2}}} = - 2\alpha .v.{v^2} = - 2\alpha {v^3}$

$	herefore $ Retardation $ = 2\alpha {v^3}$

Colum $I$	Colum $II$
$(A)$ Distance travelled in $3\,s$	$(p)$ $-20$ units
$(B)$ Displacement in $1\,s$	$(q)$ $15$ units
$(C)$ Initial acceleration	$(r)$ $25$ units
$(D)$ Velocity at $4\,s$	$(s)$ $-10$ units

The relation between time and distance is $t = \alpha {x^2} + \beta x$, where $\alpha $ and $\beta $ are constants. The retardation is

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