First slide

Rectilinear Motion

Question

The distance x covered by a body moving in a straight line in time t is given by the relation $2 x^{2} + 3 x = t$ . If v is the velocity of the body at a certain instant of time, its acceleration will be

Moderate

A

$- v^{3}$
B

$- 2 v^{3}$
C

$- 3 v^{3}$
D

$- 4 v^{3}$

Solution

Differentiating ${2x}^{2} + 3 x = t$ with respect to t we have $4 x \frac{dx}{dt} + \frac{3 dx}{dt} = 1 . . . . . (i)$
Now $\frac{dx}{dt} = v,$ Therefore, $4 xv + 3 v = 1 or 4 x + 3 = 1 / v$ .
Differentiating Eq (i) with respect to time t, we have
$4 {(\frac{dx}{dt})}^{2} + 4 x \frac{d^{2} x}{{dt}^{2}} + 3 \frac{d^{2} x}{{dt}^{2}} = 0 or 4 v^{2} + 4 xa + 3 a = 0 or a = - \frac{4 v^{2}}{4 x + 3} (ii) Where a = \frac{d^{2} x}{{dt}^{2}} is the acceleration, But 4 x + 3 = 1 / v .$
Using this in Eq.(ii) we get $a = - 4 v^{3} .$

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