Question

Матеріальна точка рухається за законом s(t) = 16 t+ 2 t3 (м). Знайдіть її швидкість і прискорення в момент часу t = 2 с. Помогите пожалуйста!!! БУДУ ООЧЕНЬ БЛАГОДАРНА!!!!

Answer 1

Объяснение:

$s(t)=16t+2t^3\ \ \ \ \ t=2 \ c.\ \ \ \ \ v(t)=?\ \ \ \ \ a(t)=?\\v(t)=s'(t)=(16t+2t^3)'=16+6t^2.\\v(2)=16+6*2^2=16+6*4=16+24=40.\\a(t)=v'(t)=(16+6t^2)'=12t.\\a(2)=12*2=24.$

Ответ: v(2)=40 м/с,  а(2)=24 м/с².

Answer 2

Problem Analysis

We are given the equation for the position of a moving object as a function of time: s(t) = 16 + t + 2t^3 (m). We need to find the velocity and acceleration of the object at t = 2 s.

Velocity Calculation

To find the velocity of the object, we need to take the derivative of the position function with respect to time. The derivative of s(t) will give us the rate of change of position with respect to time, which is velocity.

Let's calculate the derivative of s(t) with respect to t:

s(t) = 16 + t + 2t^3 (m)

Taking the derivative of s(t) with respect to t:

v(t) = d/dt (16 + t + 2t^3) = 0 + 1 + 6t^2 = 1 + 6t^2 (m/s)

Therefore, the velocity of the object at t = 2 s is:

v(2) = 1 + 6(2)^2 = 1 + 6(4) = 1 + 24 = 25 m/s

So, the velocity of the object at t = 2 s is 25 m/s.

Acceleration Calculation

To find the acceleration of the object, we need to take the derivative of the velocity function with respect to time. The derivative of v(t) will give us the rate of change of velocity with respect to time, which is acceleration.

Let's calculate the derivative of v(t) with respect to t:

v(t) = 1 + 6t^2 (m/s)

Taking the derivative of v(t) with respect to t:

a(t) = d/dt (1 + 6t^2) = 0 + 12t = 12t (m/s^2)

Therefore, the acceleration of the object at t = 2 s is:

a(2) = 12(2) = 24 m/s^2

So, the acceleration of the object at t = 2 s is 24 m/s^2.

Conclusion

The velocity of the object at t = 2 s is 25 m/s, and the acceleration of the object at t = 2 s is 24 m/s^2.