[tex]OTBET:\ t=\frac{3\pi }{4}+\pi \mathbb N[/tex].
Объяснение:
Ускорение – это вторая производная пути по времени.
[tex]\displaystyle\\\left \{ {{f(t)=sin(2t)} \atop {g(t)=2t^2+3t+1}} \right. \ \ \ \ \left \{ {{f'(t)=(sin(2t))'} \atop g'(t)=(2t^2+3t+1)'}} \right. \ \ \ \ \left \{ {{f'(t)=2cos(2t)} \atop {g'(t)=4t+3}} \right. \\\\\\\left \{ {{f''(t)=(2cos(2t))'} \atop {g''(t)=(4t+3)'} \right. \ \ \ \ \left \{ {{f''(t)=-4sin(2t)} \atop {g''(t)=4}} \right. \\\\\\f''(t)=g''(t)\ \ \ \ \Rightarrow\\\\-4sin(2t)=4\ |:(-4)\\\\sin(2t)=-1\\\\2t=\frac{3\pi }{2} +2\pi \mathbb N\ |:2\\\\t=\frac{3\pi }{4}+\pi \mathbb N[/tex]
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[tex]OTBET:\ t=\frac{3\pi }{4}+\pi \mathbb N[/tex].
Объяснение:
Ускорение – это вторая производная пути по времени.
[tex]\displaystyle\\\left \{ {{f(t)=sin(2t)} \atop {g(t)=2t^2+3t+1}} \right. \ \ \ \ \left \{ {{f'(t)=(sin(2t))'} \atop g'(t)=(2t^2+3t+1)'}} \right. \ \ \ \ \left \{ {{f'(t)=2cos(2t)} \atop {g'(t)=4t+3}} \right. \\\\\\\left \{ {{f''(t)=(2cos(2t))'} \atop {g''(t)=(4t+3)'} \right. \ \ \ \ \left \{ {{f''(t)=-4sin(2t)} \atop {g''(t)=4}} \right. \\\\\\f''(t)=g''(t)\ \ \ \ \Rightarrow\\\\-4sin(2t)=4\ |:(-4)\\\\sin(2t)=-1\\\\2t=\frac{3\pi }{2} +2\pi \mathbb N\ |:2\\\\t=\frac{3\pi }{4}+\pi \mathbb N[/tex]