Answer to Question #157154 in Mechanics | Relativity for wilder

Question

A particle starts from rest and moves in a straight line on a smooth horizontal surface. Its acceleration at time t seconds is k(4v + 1)m/s^2 , where k is a positive constant and v m/s is the speed of the particle. Given that v = (e^2 - 1)/4 when t = 1, show that v = 1/4(e^(2t) - 1)

Answer 1

A particle starts from rest  and the initial velocity is 0

$a = \large\frac{v - v_o}{t} \to v = at$

$a = k(4v+1) \large\frac{m}{s^2}$

Given that $v = \large\frac{e^2-1}{4}$ when t = 1

$v(1) = a =$ $\large\frac{e^2-1}{4}$ $= k(4v+1)$

$\large\frac{e-1}{2}*\frac{e+1}{2}$ $= k * (4v+1)$

$v = \large\frac{e^{2t} - 1}{4}$ $\to v(1) = \large\frac{e^2-1}{4}$