Question

22. Velocity of particle varies as V = x² + 2x + 5 where V is in m/s and x is in metre. Find acceleration of the particle when position of
particle x = 1 m:

(1) 4 m/s2
(2) 8 m/s2
(3) 32 m/s2
(4) 64 m/s2 ​

Answer 1

Answer:

$\boxed{\sf (3) \ 32 \ m/s^2}$

Given:

v = x² + 2x + 5

To Find:

Acceleration of particle when position of particle is x = 1 m.

Explanation:

$\sf a = \frac{dv}{dt} \\ \\ \sf \: \: \: \: = \frac{dv}{dt} \times \frac{dx}{dx} \\ \\ \sf \: \: \: \: = \frac{dx}{dt} \times \frac{dv}{dx} \\ \\ \sf \: \: \: \: = v \frac{dv}{dx} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( \frac{dx}{dt} = v) \\ \\ \sf \: \: \: \: = ( {x}^{2} + 2x + 5) \times \frac{d( {x}^{2} + 2x + 5)}{dx} \\ \\ \sf \: \: \: \: = ( {x}^{2} + 2x + 5) \times \frac{d( {x}^{2}) }{dx} + \frac{d(2x)}{dx} + \frac{d(5)}{dx} \\ \\ \sf \: \: \: \: = ( {x}^{2} + 2x + 5) \times (2x + 2) \\ \\ \sf \: \: \: \: = 2x( {x}^{2} + 2x + 5) + 2( {x}^{2} + 2x + 5)\\ \\ \sf \: \: \: \: = 2 {x}^{3} + 4 {x}^{2} + 10x + 2 {x}^{2} + 4x + 10\\ \\ \sf \: \: \: \: = 2 {x}^{3} + 6 {x}^{2} + 14x + 10$

$\sf Acceleration \ at \ x = 1: \\ \\ \sf \implies a = 2{(1)}^{3} + 6 {(1)}^{2} + 14(1) + 10 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 + 6 + 14 + 10 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 32 \: m/s^2$

So,

Acceleration of particle when position of particle is x = 1 m = 32 m/s²

Answer 2

Answer:

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