Answer to Question #165198 in Mechanics | Relativity for Dora

Question

an elementary particle is projectedin a dark hole, it is spirally and its position is defined as r=(t^2+7t)i-e^-3tj+7sin4t. find its velocityand accelerationat t=0, t=1 and at what time is particle at rest

Answer 1

$r = (t^2+7t)i - e^{-3t}j + 7sin4t$

$v = r' \space \to v = (2t+7)i +3e^{-3t}j + 28cos4t$

$v(0) = 7i+3j+28$

$v(1)= 9i+3e^{-3}j+28cos4$

$a = v' \to a = 2i - 9e^{-3t}j-112sin4t$

$a(0) = 2i - 9j$

$a(1) = 2i - 9e^{-3}j-112sin4$

time is particle at rest $v = (2t+7)i +3e^{-3t}j + 28cos4t = 0$