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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-




Question Number 184832 by Mastermind last updated on 12/Jan/23
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
$$\mathrm{Given}\:\mathrm{the}\:\mathrm{acceleration}\: \\ $$$$\mathrm{a}=−\mathrm{4sin2t},\:\mathrm{initial}\:\mathrm{velocity}\: \\ $$$$\mathrm{v}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{and}\:\mathrm{the}\:\mathrm{initial}\:\mathrm{position}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{as}\:\mathrm{s}\left(\mathrm{0}\right)=−\mathrm{3},\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{body}'\mathrm{s}\:\mathrm{position}\:\mathrm{at}\:\mathrm{time}\:\mathrm{t}. \\ $$$$ \\ $$$$\mathrm{Hi} \\ $$
Commented by manolex last updated on 12/Jan/23
don′t believe,  CHECK  you                         alcohol  s=4sin2t−3     s=sin2t−3                                  s(0)=−3  v=8cos2t            v=2cos2t                                   v(0)=−2  a=−16sin2t     a=−4sin2t  wrong                   ok
$${don}'{t}\:{believe},\:\:{CHECK} \\ $$$${you}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{alcohol} \\ $$$${s}=\mathrm{4}{sin}\mathrm{2}{t}−\mathrm{3}\:\:\:\:\:{s}={sin}\mathrm{2}{t}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{s}\left(\mathrm{0}\right)=−\mathrm{3} \\ $$$${v}=\mathrm{8}{cos}\mathrm{2}{t}\:\:\:\:\:\:\:\:\:\:\:\:{v}=\mathrm{2}{cos}\mathrm{2}{t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{v}\left(\mathrm{0}\right)=−\mathrm{2} \\ $$$${a}=−\mathrm{16}{sin}\mathrm{2}{t}\:\:\:\:\:{a}=−\mathrm{4}{sin}\mathrm{2}{t} \\ $$$${wrong}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ok} \\ $$
Answered by TheSupreme last updated on 12/Jan/23
v(t)=v(0)+∫_0 ^t adt=2+2cos(2t)  s(t)=s(0)+∫vdt=−3+2t+sin(2t)
$${v}\left({t}\right)={v}\left(\mathrm{0}\right)+\int_{\mathrm{0}} ^{{t}} {adt}=\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{2}{t}\right) \\ $$$${s}\left({t}\right)={s}\left(\mathrm{0}\right)+\int{vdt}=−\mathrm{3}+\mathrm{2}{t}+{sin}\left(\mathrm{2}{t}\right) \\ $$$$ \\ $$
Commented by Mastermind last updated on 12/Jan/23
I got s=4sin2t−3    Anyways, thank you
$$\mathrm{I}\:\mathrm{got}\:\mathrm{s}=\mathrm{4sin2t}−\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Anyways},\:\mathrm{thank}\:\mathrm{you} \\ $$
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 = (dv/(dt )) ⇒ v = 2cos(2t) + k  k = 2 − 2cos(0) = 0  ⇒ k = 0 ⇒ v = 2cos(2t)  v = (ds/dt) ⇒ s = sin(2t) + c  c = −3 − sin(0) = −3  ⇒ s = sin(2t) − 3    Alcohol
$$ \\ $$$${a}\:=\:\frac{{dv}}{{dt}\:}\:\Rightarrow\:{v}\:=\:\mathrm{2}{cos}\left(\mathrm{2}{t}\right)\:+\:{k} \\ $$$${k}\:=\:\mathrm{2}\:−\:\mathrm{2}{cos}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{k}\:=\:\mathrm{0}\:\Rightarrow\:{v}\:=\:\mathrm{2}{cos}\left(\mathrm{2}{t}\right) \\ $$$${v}\:=\:\frac{{ds}}{{dt}}\:\Rightarrow\:{s}\:=\:{sin}\left(\mathrm{2}{t}\right)\:+\:{c} \\ $$$${c}\:=\:−\mathrm{3}\:−\:{sin}\left(\mathrm{0}\right)\:=\:−\mathrm{3} \\ $$$$\Rightarrow\:{s}\:=\:{sin}\left(\mathrm{2}{t}\right)\:−\:\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Alcohol} \\ $$
Answered by Frix last updated on 12/Jan/23
s(t)=∫(∫a(t)dt)dt  ∫(∫(−4sin 2t)dt)dt=∫(2cos 2t +C_1 )dt=  =sin 2t +C_1 t+C_2   v(0)=2  2cos 0 +C_1 =2 ⇒ C_1 =0  s(0)=−3  sin 0 +C_2 =−3 ⇒ C_2 =−3  ⇒  s(t)=−3+sin 2t    Apple Juice
$${s}\left({t}\right)=\int\left(\int{a}\left({t}\right){dt}\right){dt} \\ $$$$\int\left(\int\left(−\mathrm{4sin}\:\mathrm{2}{t}\right){dt}\right){dt}=\int\left(\mathrm{2cos}\:\mathrm{2}{t}\:+{C}_{\mathrm{1}} \right){dt}= \\ $$$$=\mathrm{sin}\:\mathrm{2}{t}\:+{C}_{\mathrm{1}} {t}+{C}_{\mathrm{2}} \\ $$$${v}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$$\mathrm{2cos}\:\mathrm{0}\:+{C}_{\mathrm{1}} =\mathrm{2}\:\Rightarrow\:{C}_{\mathrm{1}} =\mathrm{0} \\ $$$${s}\left(\mathrm{0}\right)=−\mathrm{3} \\ $$$$\mathrm{sin}\:\mathrm{0}\:+{C}_{\mathrm{2}} =−\mathrm{3}\:\Rightarrow\:{C}_{\mathrm{2}} =−\mathrm{3} \\ $$$$\Rightarrow \\ $$$${s}\left({t}\right)=−\mathrm{3}+\mathrm{sin}\:\mathrm{2}{t} \\ $$$$ \\ $$$$\mathrm{Apple}\:\mathrm{Juice} \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jan/23
Appreciate your sense of humour!
$${Appreciate}\:{your}\:{sense}\:{of}\:{humour}! \\ $$
Commented by Frix last updated on 12/Jan/23
It′s better to laugh than to cry!
$$\mathrm{It}'\mathrm{s}\:\mathrm{better}\:\mathrm{to}\:\mathrm{laugh}\:\mathrm{than}\:\mathrm{to}\:\mathrm{cry}! \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jan/23
Well said sir!
Answered by a.lgnaoui last updated on 13/Jan/23
v=∫_0 ^t adt=[2cos  2t+c]_0 ^t =2sin 2t +v_0   [c=v_0 ]  v=2cos  2t+2  S=∫_0 ^t vdt=sin 2t+2t+s_0   t=0     s_0 =−3        x=sin  2t+2t−3
$${v}=\int_{\mathrm{0}} ^{{t}} {adt}=\left[\mathrm{2cos}\:\:\mathrm{2}{t}+{c}\right]_{\mathrm{0}} ^{{t}} =\mathrm{2sin}\:\mathrm{2}{t}\:+{v}_{\mathrm{0}} \:\:\left[{c}={v}_{\mathrm{0}} \right] \\ $$$${v}=\mathrm{2cos}\:\:\mathrm{2}{t}+\mathrm{2} \\ $$$${S}=\int_{\mathrm{0}} ^{{t}} {vdt}=\mathrm{sin}\:\mathrm{2}{t}+\mathrm{2}{t}+{s}_{\mathrm{0}} \\ $$$${t}=\mathrm{0}\:\:\:\:\:{s}_{\mathrm{0}} =−\mathrm{3} \\ $$$$ \\ $$$$\:\:\:\:{x}=\mathrm{sin}\:\:\mathrm{2}{t}+\mathrm{2}{t}−\mathrm{3} \\ $$$$ \\ $$

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