Question

7. The relation between time t and distance x is t = ax^2 + bx, where a and b are constants. The
acceleration is
(a) -2 abv^2
(b) -2 bv^3
(c) -2 av^3
(d) -2 av^2​

Answer 1

Answer:

$(c) \: \: \: - 2a {v}^{3}$

Explanation:

$acceleration \: a {}^{0} \: = \frac{ {d}^{2}x }{d {t}^{2} } \\ \\velocity \: v \: = \frac{dx}{dt} \\ \\ given \\ \\ t = a {x}^{2} + bx \\ \\ diff \: wrt \: t \\ \\ 1 = 2ax \times \frac{dx}{dt} + b \times \frac{dx}{dt} \\ \\ = > \frac{dx}{dt} = \frac{1}{2ax + b} = v \\ \\ diff \: wrt \: x \\ \\ = > \frac{ {d}^{2}x }{d {t}^{2} } = - \frac{1}{ {(2ax + b)}^{2} } \times 2a \times \frac{dx}{dt} \\ \\ = > {a}^{0} = - 2a {v}^{3}$

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Answer 2

Answer:

Answer:

if , t = ax^2+bx

then, differentiate both side with respect to time (t)

dt/dt = 2ax.dx/dt + b.dx/dt

1 = 2axv + bv

v(2ax + b) = 1

(2ax + b) = 1/v

Again differentiate both side with respect to time (t)

2a.dx/dt = -v^(-2)*dv/dt

2av = v^(-2)*acceleration

acceleration = -2av³

Hence,

retardation= 2av³ (ANS)

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Explanation: