Solution to problem 7.6.10 from the collection of Kepe O.E.

7.6.10 During movement, the acceleration projections of the points are determined as follows: ax = 0.8 t [m/s2], ay = 0.8 m/s2. If at the initial moment of time t0 = 0 the speed of the point is vo = 0, then it is necessary to find the tangential acceleration at the moment of time t = 2 s. The answer is 1.70.

Solution to problem 7.6.10 from the collection of Kepe O.?.

This digital product is a solution to problem 7.6.10 from the collection of Kepe O.?., associated with determining the tangential acceleration of a point at time t = 2 s during movement.

The problem is solved using the formulas: ax = 0.8 t [m/s2], ay = 0.8 m/s2, and also takes into account the initial time t0 = 0, at which the speed of the point is zero.

By purchasing this product, you receive a ready-made solution to the problem in a convenient digital format, designed in accordance with the requirements of a beautiful html design.

This digital product is a solution to problem 7.6.10 from the collection of Kepe O.?., associated with determining the tangential acceleration of a point at time t = 2 s during movement.

From the conditions of the problem it is known that the projections of the point’s acceleration are determined by the expressions ax = 0.8 t [m/s2], ay = 0.8 m/s2, and the initial speed of the point vо is equal to zero at t0 = 0.

To solve the problem, it is necessary to find the tangential acceleration of the point at time t = 2 s. The answer is 1.70.

By purchasing this product, you receive a ready-made solution to the problem in a convenient digital format, designed in accordance with the requirements of a beautiful html design.

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Problem 7.6.10 from the collection of Kepe O.?. is formulated as follows.

It is given that the projections of the acceleration of a point during movement are determined by the expressions ax = 0.8 t [m/s2], ay = 0.8 m/s2. It is necessary to find the tangential acceleration at time t = 2 s, if at t0 = 0 the speed of the point is vo = 0.

To solve the problem, you need to use the formula for calculating tangential acceleration:

at = √(ah² + au²)

where at is the tangential acceleration, akh and ay are the projections of the acceleration of the point along the x and y axes, respectively.

Substituting the values ​​of ax and ay from the problem conditions, we get:

at = √(0,8²·2² + 0,8²) ≈ 1,70 [м/с²]

Thus, the tangential acceleration at time t = 2 s is 1.70 m/s².

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