Question

Найдите скорость и ускорение точки в момент t0, если: x(t)=2t^2+t-4, t0=4 Туундунун механикалык мааниси эмнеде?

Answer 1

Ответ:

Объяснение:

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Answer 2

Finding the Velocity and Acceleration of a Point at t0

To find the velocity and acceleration of a point at a specific moment t0, we need to differentiate the position function with respect to time.

Given the position function x(t) = 2t^2 + t - 4 and t0 = 4, we can find the velocity and acceleration at t0 by differentiating x(t) with respect to t.

1. Velocity: The velocity of a point is the derivative of its position function with respect to time. Taking the derivative of x(t) will give us the velocity function v(t).

Differentiating x(t) = 2t^2 + t - 4 with respect to t, we get: x'(t) = 4t + 1

To find the velocity at t0 = 4, we substitute t = 4 into the velocity function: v(t0) = 4(4) + 1 = 16 + 1 = 17

Therefore, the velocity of the point at t0 = 4 is 17 units per time.

2. Acceleration: The acceleration of a point is the derivative of its velocity function with respect to time. Taking the derivative of v(t) will give us the acceleration function a(t).

Differentiating v(t) = 4t + 1 with respect to t, we get: v'(t) = 4

The acceleration is constant, and it does not depend on time.

Therefore, the acceleration of the point is 4 units per time squared.

To summarize: - The velocity of the point at t0 = 4 is 17 units per time. - The acceleration of the point is 4 units per time squared.

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