Solution to problem 7.8.3 from the collection of Kepe O.E.

Let's solve the problem:

Hopefully:

Tangential acceleration of point a_etc = 1.4 m/s²

Total acceleration of point a = 2.6 m/s²

Find:

Normal acceleration of point a_n

Answer:

It is known that the total acceleration of a point is the vector sum of the tangential and normal accelerations:

a = a_etc + a_n

Since the vector product of the tangent and normal accelerations is zero, the accelerations are perpendicular to each other:

a_etc·a_n = 0

It follows that:

a_n = √(a)² - (a_etc)²

a_n = √(2.6 m/s²)² - (1.4 m/s²)² = 2.19 m/s²

Answer: a_n = 2.19 m/s².

Cargo code: 7.8.3-KO

Product name: Solution to problem 7.8.3 from the collection of Kepe O.?.

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The product is the solution to problem 7.8.3 from the collection of problems "Kepe O.?".

In this problem, it is necessary to determine the normal acceleration of a point moving along a curved path at the time when its total acceleration is 2.6 m/s2 and its tangential acceleration is 1.4 m/s2.

To solve the problem, it is necessary to use a formula to calculate the total acceleration of a point, which is represented as a vector sum of the tangential and normal accelerations. Knowing the tangential acceleration and total acceleration, you can find the normal acceleration.

After substituting known values ​​into the formula and solving mathematical expressions, we get the answer: the normal acceleration of a point at the moment of time when its total acceleration a = 2.6 m/s2 is equal to 2.19 m/s2.

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