Question

S(t)= 1/6t^3+t^2+4-1 (м). t=2c

Answer 1

Объяснение:

$S(t)=\frac{1}{6}t^3+t^2+4t-1\ \ \ \ t=2 c\ \ \ \ a(2)= ?\\v(t)=S'(t)=(\frac{1}{6}t^3+t^2+4t-1)'=\frac{3}{6} t^2+2t+4=\frac{t^2}{2}+2t+4.\\a(t)=v'(t)=(\frac{t^2}{2}+2t+4)'=\frac{2t}{2} +2=t+2.\\a(2)=2+2=4.$

Ответ: a(2)=4 м/с².

Answer 2

To find the position of an object at a specific time, you can simply plug in the given time value into the position equation. In this case, you're given:

S(t) = (1/6)t^3 + t^2 + 4 - 1 (in meters)

And t = 2 seconds.

So, plug in t = 2 into the equation:

S(2) = (1/6)(2^3) + 2^2 + 4 - 1

S(2) = (1/6)(8) + 4 + 4 - 1

Now, calculate the values:

S(2) = (4/3) + 4 + 4 - 1

S(2) = (4/3) + 12 - 1

To simplify, find a common denominator:

S(2) = (4/3) + (36/3) - (3/3)

Now, add the fractions:

S(2) = (4 + 36 - 3)/3

S(2) = 37/3 meters

So, at t = 2 seconds, the position of the object is 37/3 meters.