Given the acceleration a=−4sin2t, initial velocity v(0)=2, and the initial position of the body as s(0)=−3, find the body′s position at time t. Hi


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Question Number 184832 by Mastermind last updated on 12/Jan/23

$Given the acceleration$ $a = - 4 \sin 2 t, initial velocity$ $v (0) = 2, and the initial position$ $of the body as s (0) = - 3, find the$ ${body}^{'} s position at time t .$ $Hi$
Commented by manolex last updated on 12/Jan/23

${d o n}^{'} t b e l i e v e, C H E C K$ $y o u a l c o h o l$ $s = 4 s i n 2 t - 3 s = s i n 2 t - 3$ $s (0) = - 3$ $v = 8 c o s 2 t v = 2 c o s 2 t$ $v (0) = - 2$ $a = - 16 s i n 2 t a = - 4 s i n 2 t$ $w r o n g o k$
	Answered by TheSupreme last updated on 12/Jan/23

	$v (t) = v (0) + \int_{0}^{t} a d t = 2 + 2 c o s (2 t)$ $s (t) = s (0) + \int v d t = - 3 + 2 t + s i n (2 t)$
	Commented by Mastermind last updated on 12/Jan/23

	$I got s = 4 \sin 2 t - 3$ $Anyways, thank you$
	Commented by JDamian last updated on 12/Jan/23
	which is obviously wrong. Have you tried to get s'' and compare with a?
	Answered by alcohol last updated on 12/Jan/23

	$a = \frac{d v}{d t} \Rightarrow v = 2 c o s (2 t) + k$ $k = 2 - 2 c o s (0) = 0$ $\Rightarrow k = 0 \Rightarrow v = 2 c o s (2 t)$ $v = \frac{d s}{d t} \Rightarrow s = s i n (2 t) + c$ $c = - 3 - s i n (0) = - 3$ $\Rightarrow s = s i n (2 t) - 3$ $Alcohol$
	Answered by Frix last updated on 12/Jan/23

	$s (t) = \int (\int a (t) d t) d t$ $\int (\int (- 4 \sin 2 t) d t) d t = \int (2 \cos 2 t + C_{1}) d t =$ $= \sin 2 t + C_{1} t + C_{2}$ $v (0) = 2$ $2 \cos 0 + C_{1} = 2 \Rightarrow C_{1} = 0$ $s (0) = - 3$ $\sin 0 + C_{2} = - 3 \Rightarrow C_{2} = - 3$ $\Rightarrow$ $s (t) = - 3 + \sin 2 t$ $Apple Juice$
	Commented by Rasheed.Sindhi last updated on 12/Jan/23

	$A p p r e c i a t e y o u r s e n s e o f h u m o u r!$
	Commented by Frix last updated on 12/Jan/23

	${It}^{'} s better to laugh than to cry!$
	Commented by Rasheed.Sindhi last updated on 12/Jan/23
	Well said sir!
	Answered by a.lgnaoui last updated on 13/Jan/23

	$v = \int_{0}^{t} a d t = {[2 \cos 2 t + c]}_{0}^{t} = 2 \sin 2 t + v_{0} [c = v_{0}]$ $v = 2 \cos 2 t + 2$ $S = \int_{0}^{t} v d t = \sin 2 t + 2 t + s_{0}$ $t = 0 s_{0} = - 3$ $x = \sin 2 t + 2 t - 3$
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