υ⃗ (t)=A⋅(tτ)3⋅i⃗ +B⋅(tτ)5⋅j⃗ ,υ⃗ (t)=3⋅(t1)3⋅i⃗ +4⋅(t1)5⋅j⃗ ,υ⃗ (t)=3⋅t3⋅i⃗ +4⋅t5⋅j⃗ (1).x=x0+∫0t3⋅t3dt=x0+3⋅14⋅t4∣∣∣t0=x0+3⋅t44,x0=0,x=3⋅t44(2).y=y0+∫0t4⋅t5dt=y0+4⋅16⋅t6∣∣∣t0=y0+4⋅t66,y0=0,y=4⋅t66=2⋅t63(3).s=(3⋅t44)2+(2⋅t63)2−−−−−−−−−−−−−−−√(4).s=(3⋅144)2+(2⋅163)2−−−−−−−−−−−−−−−−√=1.
ответ: д) 1,00 м.
Объяснение:
https://www.alsak.ru/smf/index.php?topic=26088.0
υ⃗ (t)=A⋅(tτ)3⋅i⃗ +B⋅(tτ)5⋅j⃗ ,υ⃗ (t)=3⋅(t1)3⋅i⃗ +4⋅(t1)5⋅j⃗ ,υ⃗ (t)=3⋅t3⋅i⃗ +4⋅t5⋅j⃗ (1).x=x0+∫0t3⋅t3dt=x0+3⋅14⋅t4∣∣∣t0=x0+3⋅t44,x0=0,x=3⋅t44(2).y=y0+∫0t4⋅t5dt=y0+4⋅16⋅t6∣∣∣t0=y0+4⋅t66,y0=0,y=4⋅t66=2⋅t63(3).s=(3⋅t44)2+(2⋅t63)2−−−−−−−−−−−−−−−√(4).s=(3⋅144)2+(2⋅163)2−−−−−−−−−−−−−−−−√=1.
ответ: д) 1,00 м.
Объяснение:
https://www.alsak.ru/smf/index.php?topic=26088.0