If an object covering distances is in direct proportion to the square of the time elapsed, then the acceleration is

Increasing

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Decreasing

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Constant

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None of these

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Solution

The correct option is C Constant
$x = k t^{2}$
$v e l o c i t y = \frac{d x}{d t} = \frac{d (k t^{2})}{d t} = 2 k t$
$a c c e l e r a t i o n = \frac{v e l o c i t y}{d t} = \frac{d (2 k t)}{d t} = 2 k$ where k is constant.

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If an object covering distances is in direct proportion to the square of the time elapsed, then the acceleration is

The correct option is C Constantx=kt2velocity=dxdt=d(kt2)dt=2ktacceleration=velocitydt=d(2kt)dt=2kwhere k is constant.

The correct option is C Constant
$x = k t^{2}$
$v e l o c i t y = \frac{d x}{d t} = \frac{d (k t^{2})}{d t} = 2 k t$
$a c c e l e r a t i o n = \frac{v e l o c i t y}{d t} = \frac{d (2 k t)}{d t} = 2 k$ where k is constant.