Question

Материальная точка движется прямолинейно. Уравнение имеет вид x = At + Bt^2 , где A = 3 м/с; B = 0,06 м/с^3 ; . Найти скорость V и ускорение точки в моменты времени t1 = 0 и t2 = 3 c. Каковы средние значения скорости и ускорения за первые 3 с. движения?

Answer 1

X=3t+0,06t^2; v(t)=x'=3+0,12t, v(0)=3 м/с v(3)=3+0,12*3=3,36 м/с, а(t)=v'(t)=0,12 м/с2- движение с постоянным ускорением.Средняя скорость за 3с Vc=S/t=(3*3+0,06*3^2)/3=9,54/3=3,18 м/с

Answer 2

Given Information

We are given that the material point moves in a straight line and its equation of motion is given by x = At + Bt^2, where A = 3 m/s and B = 0.06 m/s^3. We need to find the velocity V and acceleration of the point at the moments in time t1 = 0 and t2 = 3 seconds. We also need to calculate the average values of velocity and acceleration for the first 3 seconds of motion.

Finding Velocity and Acceleration at t1 = 0

To find the velocity at t1 = 0, we need to differentiate the equation of motion with respect to time t. The derivative of x with respect to t gives us the velocity V.

Differentiating x = At + Bt^2 with respect to t: dx/dt = A + 2Bt

Substituting the given values A = 3 m/s and B = 0.06 m/s^3, we can find the velocity V at t1 = 0.

V = dx/dt = A + 2Bt V = 3 + 2(0.06)(0) = 3 m/s

To find the acceleration at t1 = 0, we need to differentiate the velocity V with respect to time t. The derivative of V with respect to t gives us the acceleration.

Differentiating V = A + 2Bt with respect to t: dV/dt = 2B

Substituting the given value B = 0.06 m/s^3, we can find the acceleration at t1 = 0.

a = dV/dt = 2B a = 2(0.06) = 0.12 m/s^2

Therefore, at t1 = 0, the velocity V is 3 m/s and the acceleration a is 0.12 m/s^2.

Finding Velocity and Acceleration at t2 = 3

To find the velocity at t2 = 3, we can substitute t = 3 into the equation of motion x = At + Bt^2 and solve for x. Then, we can differentiate x with respect to t to find the velocity V.

Substituting t = 3 into x = At + Bt^2: x = A(3) + B(3)^2 x = 3A + 9B

Differentiating x = 3A + 9B with respect to t: dx/dt = d(3A + 9B)/dt dx/dt = 0 + 0

Since the derivative of a constant is zero, the velocity V at t2 = 3 is zero.

To find the acceleration at t2 = 3, we need to differentiate the velocity V with respect to time t. The derivative of V with respect to t gives us the acceleration.

Differentiating V = 0 with respect to t: dV/dt = 0

Since the derivative of a constant is zero, the acceleration a at t2 = 3 is zero.

Therefore, at t2 = 3, the velocity V is 0 m/s and the acceleration a is 0 m/s^2.

Average Values of Velocity and Acceleration for the First 3 Seconds

To find the average value of velocity for the first 3 seconds, we can calculate the change in position and divide it by the change in time.

The change in position Δx for the first 3 seconds can be found by substituting t = 3 into the equation of motion x = At + Bt^2 and subtracting the initial position x = 0.

Substituting t = 3 into x = At + Bt^2: x = A(3) + B(3)^2 x = 3A + 9B

Since the initial position x = 0, the change in position Δx is: Δx = x - x0 = 3A + 9B - 0 = 3A + 9B

The change in time Δt for the first 3 seconds is 3 seconds.

Therefore, the average value of velocity Vavg for the first 3 seconds is: Vavg = Δx/Δt = (3A + 9B)/3 = A + 3B

Substituting the given values A = 3 m/s and B = 0.06 m/s^3, we can find the average value of velocity Vavg for the first 3 seconds.

Vavg = 3 + 3(0.06) = 3.18 m/s

To find the average value of acceleration for the first 3 seconds, we can calculate the change in velocity and divide it by the change in time.

The change in velocity ΔV for the first 3 seconds is V - V0, where V is the velocity at t2 = 3 and V0 is the velocity at t1 = 0.

Since V = 0 m/s and V0 = 3 m/s, the change in velocity ΔV is: ΔV = V - V0 = 0 - 3 = -3 m/s

Therefore, the average value of acceleration aavg for the first 3 seconds is: aavg = ΔV/Δt = -3/3 = -1 m/s^2

Therefore, the average value of velocity Vavg for the first 3 seconds is 3.18 m/s and the average value of acceleration aavg for the first 3 seconds is -1 m/s^2.

Note: The given equation x = At + Bt^2 does not match any of the search result snippets provided. However, the solution provided above follows the standard procedure for finding velocity and acceleration from a given equation of motion.