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OMNOM
@OMNOM
July 2022
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ПОМОГИТЕ, ПОЖАЛУЙСТА) СРОЧНОО!! Движение точки задано уравнением х = 12t - 2t^2. Определите среднюю скорость движения точки в интервале времени от t1 = 1с до t2 = 4 с.
Ответ: 2 м/c
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radikkhammatov
Х=12t-4t^2/2 x=ut+at/2
u=12
a= -4
s1=12-2=10 t1=1c
ucp1=10/1=10м/c^2
s2=48-32=16 t=4c
ucp2=16/4=4м/с^2
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Answers & Comments
u=12
a= -4
s1=12-2=10 t1=1c
ucp1=10/1=10м/c^2
s2=48-32=16 t=4c
ucp2=16/4=4м/с^2