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Question Number 15039 by Tinkutara last updated on 07/Jun/17
The acceleration of an object is given by a(t) = cos(πt) ms^(−2) and its velocity at time t = 0 is (1/(2π)) m/s at origin. Its velocity at t = (3/2) s is? The object′s position at t = (3/2) s is?
$$\mathrm{The}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{an}\:\mathrm{object}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by}\:{a}\left({t}\right)\:=\:\mathrm{cos}\left(\pi{t}\right)\:\mathrm{ms}^{−\mathrm{2}} \:\mathrm{and}\:\mathrm{its}\:\mathrm{velocity} \\ $$$$\mathrm{at}\:\mathrm{time}\:{t}\:=\:\mathrm{0}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\mathrm{m}/\mathrm{s}\:\mathrm{at}\:\mathrm{origin}.\:\mathrm{Its} \\ $$$$\mathrm{velocity}\:\mathrm{at}\:{t}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{s}\:\mathrm{is}? \\ $$$$\mathrm{The}\:\mathrm{object}'\mathrm{s}\:\mathrm{position}\:\mathrm{at}\:{t}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{s}\:\mathrm{is}? \\ $$
Answered by ajfour last updated on 07/Jun/17
v−(1/(2π))=((sin πt)/π) v∣_(t=3/2) =(1/(2π))−(1/π)=−(1/(2π)) m/s x=(t/(2π))−(((cos πt−1))/π^2 ) x∣_(t=3/2) =((3/(4π))+(1/π^2 )) m .
$${v}−\frac{\mathrm{1}}{\mathrm{2}\pi}=\frac{\mathrm{sin}\:\pi{t}}{\pi} \\ $$$${v}\mid_{{t}=\mathrm{3}/\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}\pi}−\frac{\mathrm{1}}{\pi}=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:{m}/{s} \\ $$$${x}=\frac{{t}}{\mathrm{2}\pi}−\frac{\left(\mathrm{cos}\:\pi{t}−\mathrm{1}\right)}{\pi^{\mathrm{2}} } \\ $$$${x}\mid_{{t}=\mathrm{3}/\mathrm{2}} =\left(\frac{\mathrm{3}}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\right)\:{m}\:. \\ $$
Commented by Tinkutara last updated on 07/Jun/17
But answer given is (i): (3/(2π)) m/s (ii): (3/(4π)) + (1/π^2 ) m
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{given}\:\mathrm{is}\:\left(\mathrm{i}\right):\:\frac{\mathrm{3}}{\mathrm{2}\pi}\:\mathrm{m}/\mathrm{s} \\ $$$$\left(\mathrm{ii}\right):\:\frac{\mathrm{3}}{\mathrm{4}\pi}\:+\:\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:\mathrm{m} \\ $$
Commented by mrW1 last updated on 07/Jun/17
the answer given in your book is wrong.
$$\mathrm{the}\:\mathrm{answer}\:\mathrm{given}\:\mathrm{in}\:\mathrm{your}\:\mathrm{book}\:\mathrm{is}\:\mathrm{wrong}. \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by mrW1 last updated on 07/Jun/17
v(t)=∫a(t)dt=∫cos (πt)dt=((sin (πt))/π)+C since v=(1/(2π)) at t=0 ⇒C=(1/(2π)) ⇒v(t)=((sin (πt))/π)+(1/(2π)) x(t)=∫v(t)dt=−((cos (πt))/π^2 )+(t/(2π))+C since x=0 at t=0⇒C=(1/π^2 ) ⇒x(t)=((1−cos (πt))/π^2 )+(t/(2π)) at t=(3/2) v=−(1/π)+(1/(2π))=−(1/(2π)) x=(1/π^2 )+(3/(4π))
$$\mathrm{v}\left(\mathrm{t}\right)=\int\mathrm{a}\left(\mathrm{t}\right)\mathrm{dt}=\int\mathrm{cos}\:\left(\pi\mathrm{t}\right)\mathrm{dt}=\frac{\mathrm{sin}\:\left(\pi\mathrm{t}\right)}{\pi}+\mathrm{C} \\ $$$$\mathrm{since}\:\mathrm{v}=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\mathrm{at}\:\mathrm{t}=\mathrm{0}\:\Rightarrow\mathrm{C}=\frac{\mathrm{1}}{\mathrm{2}\pi} \\ $$$$\Rightarrow\mathrm{v}\left(\mathrm{t}\right)=\frac{\mathrm{sin}\:\left(\pi\mathrm{t}\right)}{\pi}+\frac{\mathrm{1}}{\mathrm{2}\pi} \\ $$$$\mathrm{x}\left(\mathrm{t}\right)=\int\mathrm{v}\left(\mathrm{t}\right)\mathrm{dt}=−\frac{\mathrm{cos}\:\left(\pi\mathrm{t}\right)}{\pi^{\mathrm{2}} }+\frac{\mathrm{t}}{\mathrm{2}\pi}+\mathrm{C} \\ $$$$\mathrm{since}\:\mathrm{x}=\mathrm{0}\:\mathrm{at}\:\mathrm{t}=\mathrm{0}\Rightarrow\mathrm{C}=\frac{\mathrm{1}}{\pi^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{x}\left(\mathrm{t}\right)=\frac{\mathrm{1}−\mathrm{cos}\:\left(\pi\mathrm{t}\right)}{\pi^{\mathrm{2}} }+\frac{\mathrm{t}}{\mathrm{2}\pi} \\ $$$$ \\ $$$$\mathrm{at}\:\mathrm{t}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{v}=−\frac{\mathrm{1}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}\pi}=−\frac{\mathrm{1}}{\mathrm{2}\pi} \\ $$$$\mathrm{x}=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}\pi} \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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