Question

please. answer above question​

Answer 1

Answer:

${v}^{2} = 2x(x + 1)$

Explanation:

$given \: \: a = 2x + 1 \\ we \: \: know \: that \: \: a = \frac{dv}{dx } \\ a = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx} \\ 2x + 1 = v\frac{dv}{dx} \\ (2x + 1)dx = vdv \\ \int(2x + 1)dx = \int vdv \\ 2 \times \frac{ {x}^{2} }{2} + x = \frac{ {v}^{2} }{2} + c \\ {x}^{2} + x = \frac{ {v}^{2} }{2} + c \\ A/Q, \: \: \: \: v = 0 \: \: at \: \: \: x = 0 \\ then \: \: \: \\ {0}^{2} + 0 = \frac{ {0}^{2} }{2} + c \\ c = 0 \\so, velocity-displacement\:\: equation \\ {x}^{2} + x = \frac{ {v}^{2} }{2} \\ {v}^{2} = 2x(x + 1)$