Mng ơi giúp mik với ạ .mik cảm ơn nhìu

Câu 15: 

Ta có: 

\(\begin{array}{l}
v = 5\pi \cos \left( {\pi t - \dfrac{\pi }{2}} \right)\\
 \Rightarrow x = 5\cos \left( {\pi t - \pi } \right)\\
 \Rightarrow a = 5{\pi ^2}\cos \left( {\pi t + \pi } \right)
\end{array}\)

Tại thời điểm \(t = \dfrac{1}{4}s\):

\(\left\{ \begin{array}{l}
x = 5\cos \left( {\dfrac{1}{4}\pi  - \pi } \right) =  - 2,5\sqrt 2 cm\\
v = 5\pi \cos \left( {\dfrac{1}{4}.\pi  - \dfrac{\pi }{2}} \right) = 2,5\pi \sqrt 2 \left( {cm/s} \right)\\
a = 5{\pi ^2}\cos \left( {\dfrac{1}{4}\pi  + \pi } \right) =  - 2,5{\pi ^2}\sqrt 2 \left( {cm/{s^2}} \right)
\end{array} \right.\)

Câu 16:

Ta có:

\(\begin{array}{l}
a = 5{\pi ^2}\cos \left( {\pi t - \dfrac{\pi }{4}} \right)\\
 \Rightarrow v = 5\pi \cos \left( {\pi t - \dfrac{{3\pi }}{4}} \right)\\
 \Rightarrow x = 5\cos \left( {\pi t - \dfrac{{5\pi }}{4}} \right)
\end{array}\)

Tại thời điểm \(t = \dfrac{1}{2}s\):

\(\left\{ \begin{array}{l}
x = 5\cos \left( {\dfrac{\pi }{2} - \dfrac{{5\pi }}{4}} \right) =  - 2,5\sqrt 2 \left( {cm} \right)\\
v = 5\pi \cos \left( {\dfrac{\pi }{2} - \dfrac{{3\pi }}{4}} \right) = 2,5\pi \sqrt 2 \left( {cm/s} \right)\\
a = 5{\pi ^2}\cos \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right) = 2,5{\pi ^2}\sqrt 2 \left( {cm/{s^2}} \right)
\end{array} \right.\)

Đáp án+Giải thích các bước giải:

$x=Acos(ωt+φ)$
$v=Aωcos(ωt+φ+\frac{\pi}{2})$ 
$a=Acos(ωt+φ+\pi)$ 
Thời điểm ban đầu: $t=0$

$a)$ 
$v=5\pi.cos(\pi.t-\frac{\pi}{2})$ 
$+) t=0→v=5\pi.cos(\pi.0-\frac{\pi}{2})=0 (cm/s)$
$+) t=\frac{1}{4}→v=5\pi.cos(\pi.\frac{1}{4}-\frac{\pi}{2})≈11,1 (cm/s)$

$x=5cos(\pi.t-\frac{\pi}{2}-\frac{\pi}{2})$ 
$=5cos(\pi.t-\pi)$ 
$+) t=0→x=5cos(\pi.0-\pi)=-5 (cm)$
$+) t=\frac{1}{4}→x=5cos(\pi.\frac{1}{4}-\pi)≈-3,5 (cm)$

$a=5\pi^{2}.cos(\pi.t-\pi+\pi)$ 
$=5\pi^{2}.cos(\pi.t)$ 
$+) t=0→a=5\pi^{2}.cos(\pi.0)≈49,35 (cm^{2}/s)$
$+) t=\frac{1}{4}→a=5\pi^{2}.cos(\pi.\frac{1}{4})≈34,89 (cm^{2}/s)$

$b)$
$a=5\pi^{2}.cos(\pi.t-\frac{\pi}{4})$ 
$+) t=0→a=5\pi^{2}.cos(\pi.0-\frac{\pi}{4})≈34,89 (cm^{2}/s)$
$+) t=\frac{1}{2}→a=5\pi^{2}.cos(\pi.\frac{1}{2}-\frac{\pi}{4})≈34,89 (cm^{2}/s)$

$x=5cos(\pi.t-\frac{\pi}{4}-\pi)$ 
$=5cos(\pi.t-\frac{5\pi}{4})$ 
$+) t=0→x=5cos(\pi.0-\frac{5\pi}{4})≈-3,5 (cm)$
$+) t=\frac{1}{2}→x=5cos(\pi.\frac{1}{2}-\frac{5\pi}{4})≈-3,5 (cm)$ 

$v=5\pi.cos(\pi.t-\frac{\pi}{4}-\frac{\pi}{2})$ 
$=5\pi.cos(\pi.t-\frac{3\pi}{4})$ 
$+) t=0→v=5\pi.cos(\pi.0-\frac{3\pi}{4})≈-11,1 (cm/s)$
$+) t=\frac{1}{2}→v=5\pi.cos(\pi.\frac{1}{2}-\frac{3\pi}{4})≈11,1 (cm/s)$