Solution to problem 7.8.15 from the collection of Kepe O.E.

7.8.15

There is a speed graph v = v(t) motion of a point along a circle of radius 8 m. It is necessary to find the total acceleration at the moment of time t = 4 s.

Answer: 2.24.

To solve the problem, you need to use the formula to calculate the total acceleration of a point moving in a circle: a_full = √(a_tight² + a_Work²), Where a_tight = dv/dt - tangential acceleration, a_Work = v²/R - radial acceleration, v - point speed, R - radius of the circle.

On the graph you can determine the speed value v at a point in time t = 4 s. To do this, you need to find a point on the graph corresponding to a given time value. After this, the tangential acceleration can be calculated a_tight as the derivative of speed with respect to time at a given point in time. Radial acceleration a_Work can be calculated knowing the speed value at a given time and the radius of the circle.

Substituting the found values ​​into the formula for total acceleration, we get the answer 2.24.

Solution to problem 7.8.15 from the collection of Kepe O..

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Problem 7.8.15 from the collection of Kepe O.?. consists of determining the total acceleration of a point moving in a circle of radius 8 meters at time t = 4 seconds. To solve the problem, you need to know the speed of the point at time t = 4 seconds, as well as the radius of the circle. The speed can be determined from the speed graph v = v(t) presented in the problem statement, and the radius of the circle is specified explicitly.

The total acceleration of a point can be determined by the formula:

a = √(a_t^2 + a_n^2)

where a_t is tangential acceleration, a_n is normal acceleration. Tangential acceleration is defined as the derivative of velocity with respect to time multiplied by the radius of the circle:

a_t = dv/dt * r

Normal acceleration is defined as the speed squared divided by the radius of the circle:

a_n = v^2 / r

Substituting the known values, we get:

a_t = 8 * π/4 = 2π м/с^2

v = 4 m/s

r = 8 m

a_n = v^2 / r = 4^2 / 8 = 2 м/с^2

a = √(a_t^2 + a_n^2) = √(4π^2 + 4) ≈ 2.24 м/с^2

Thus, the total acceleration of the point at time t = 4 seconds is 2.24 m/s^2.

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