Question

Материальная точка движется прямолинейно. Уравнение движения имеет вид s = 2t + 0,04t3 (расстояние – в метрах, время – в секундах). Найти скорость и ускорение точки в моменты времени t1 = 0 и t2 = 5 с. Каковы средние значения скорости и ускорения за первые 5 с движения?

Answer 1

Скорость движения  - это производная от  пути по времени, то есть v(t)=2+0.12*t^2, Ускорение - производная от скорости по времени a(t)=0.24*t. v(t=0)=2 м/с v(t=5)=5 м/с. a(t=0)=0 м/с^2, a(t=5)=0.24*5=1.2 м/с^2. За 5 секунд тело прошло s=2*5+0.04*5^3=15 м, то есть средняя скорость за 5 секунд составляет Vср=15/5=3 м/с. Ускорение меняется по линейному закону, поэтому среднее возьмём как среднее арифметическое aср=(0+1,2)/2=0,6 м/с^2.

Answer 2

According to the given equation of motion, s = 2t + 0.04t^3, where s represents the distance in meters and t represents the time in seconds.

To find the velocity and acceleration of the point at the moments t1 = 0 and t2 = 5 seconds, we need to differentiate the equation of motion with respect to time.

Velocity:

The velocity of an object is the rate of change of its position with respect to time. It can be found by differentiating the equation of motion with respect to time.

Differentiating the equation s = 2t + 0.04t^3 with respect to t, we get:

v = ds/dt = 2 + 0.12t^2 [[1]]

To find the velocity at t1 = 0 and t2 = 5 seconds, we substitute these values into the equation for velocity:

At t1 = 0 seconds: v1 = 2 + 0.12(0)^2 = 2 m/s

At t2 = 5 seconds: v2 = 2 + 0.12(5)^2 = 2 + 0.12(25) = 2 + 3 = 5 m/s

Acceleration:

The acceleration of an object is the rate of change of its velocity with respect to time. It can be found by differentiating the equation for velocity with respect to time.

Differentiating the equation v = 2 + 0.12t^2 with respect to t, we get:

a = dv/dt = 0.24t [[2]]

To find the acceleration at t1 = 0 and t2 = 5 seconds, we substitute these values into the equation for acceleration:

At t1 = 0 seconds: a1 = 0.24(0) = 0 m/s^2

At t2 = 5 seconds: a2 = 0.24(5) = 1.2 m/s^2

Average Velocity and Acceleration:

To find the average velocity and acceleration for the first 5 seconds of motion, we need to calculate the total change in position and velocity over that time interval.

The change in position over the first 5 seconds can be found by substituting t2 = 5 into the equation for position:

s2 = 2(5) + 0.04(5)^3 = 10 + 0.04(125) = 10 + 5 = 15 m

The change in velocity over the first 5 seconds can be found by subtracting the initial velocity from the final velocity:

Δv = v2 - v1 = 5 - 2 = 3 m/s

The average velocity over the first 5 seconds is given by the ratio of the change in position to the time interval:

average velocity = Δs / Δt = (s2 - s1) / (t2 - t1) = (15 - 0) / (5 - 0) = 3 m/s

The average acceleration over the first 5 seconds is given by the ratio of the change in velocity to the time interval:

average acceleration = Δv / Δt = (v2 - v1) / (t2 - t1) = (5 - 2) / (5 - 0) = 0.6 m/s^2

Therefore, the velocity of the point at t1 = 0 seconds is 2 m/s, and at t2 = 5 seconds is 5 m/s. The acceleration of the point at t1 = 0 seconds is 0 m/s^2, and at t2 = 5 seconds is 1.2 m/s^2. The average velocity over the first 5 seconds is 3 m/s, and the average acceleration over the first 5 seconds is 0.6 m/s^2.

Please let me know if there's anything else I can help you with!