MCQ
The relation between time and distance is $t = \alpha {x^2} + \beta x$, where $\alpha $ and $\beta $ are constants. The retardation is
- ✓$2\alpha {v^3}$
- B$2\beta {v^3}$
- C$2\alpha \beta {v^3}$
- D$2{\beta ^2}{v^3}$
✓
Answer
Correct option: A.
$2\alpha {v^3}$
a
(a) $\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}$
(a) $\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}$
$a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}$
$a = v\frac{{dv}}{{dx}} = \frac{{ - v.2\alpha }}{{{{(2\alpha x + \beta )}^2}}} = - 2\alpha .v.{v^2} = - 2\alpha {v^3}$
$\therefore $ Retardation $ = 2\alpha {v^3}$
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