Solution to problem 7.8.13 from the collection of Kepe O.E.

7.8.13 A point moves along a circle of radius r = 6 m with a speed v = 3t. Determine the angle in degrees between the acceleration and velocity of the point at time t = 1 s. (Answer 26.6)

Let's consider the movement of a point along a circle of radius $r=6$ meters. It is known that its speed is determined by the formula $v=3t$, where $t$ is the time of movement. It is necessary to find the angle between the acceleration and velocity vectors of a point at time $t=1$ second.

Solution: The speed of a point can be expressed through the angular speed $\omega$ and the radius of the circle $r$: $$v = r\omega.$$ Thus, the angular speed is equal to $\omega = \frac{v}{r} = \ frac{3t}{r}.$

The acceleration of a point in a given motion is constantly directed towards the center of the circle and is determined by the formula $a=\frac{v^2}{r}$. Thus, the acceleration of the point is equal to $a=\frac{(3t)^2}{r}=\frac{9t^2}{r}$.

At the moment of time $t=1$ second, the angular velocity is equal to $\omega=\frac{3}{6}=0.5$ rad/s, and the acceleration is equal to $a=\frac{9}{6}=1.5$ m/ c$^2$. The angle between the acceleration and velocity vectors can be found using the formula: $$\cos\alpha=\frac{\vec{v}\cdot\vec{a}}{|\vec{v}|\cdot|\vec{a }|}.$$

Substituting the values ​​into this formula, we get: $$\cos\alpha=\frac{(3\cdot1)\cdot(9/6)}{(3\cdot1)\cdot\sqrt{(9/6)^2+ (3/2)^2}}\approx0.453,$$ whence $\alpha\approx26.6$ degrees. Thus, the desired angle is 26.6 degrees.

This digital product is a solution to problem 7.8.13 from a collection of problems in physics, authored by O.?. Kepe. The product is an electronic file containing a detailed and understandable solution to this problem, which can be used to prepare for exams, independently study physics and solve similar problems.

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Description of the product: this digital product is a solution to problem 7.8.13 from the collection of problems in physics, authored by O.?. Kepe. The problem is to move a point along a circle of radius 6 m, with a speed v = 3t, where t is the time of movement. It is necessary to find the angle in degrees between the acceleration vector and the velocity vector of the point at time t = 1 s.

A digital product is an electronic file in a convenient and understandable html format containing a detailed and step-by-step solution to this problem. The file contains detailed calculations and explanations of each step in solving the problem.

This product can be used to prepare for exams, independently study physics and solve similar problems. It is an excellent choice for anyone who wants to improve their knowledge of physics and successfully cope with exam preparation.

Answer to problem 7.8.13 from the collection of Kepe O.?. equal to 26.6 degrees.

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The product is the solution to problem 7.8.13 from the collection of Kepe O.?. The problem is formulated as follows: on a circle of radius r = 6 m, a point moves with a speed v = 3t. It is necessary to find the angle between the acceleration and speed of the point at time t = 1 s. The answer to the problem is 26.6 degrees.

To solve the problem, it is necessary to determine the radius vector of the point at time t = 1 s, as well as its speed and acceleration. The radius vector of the point will be equal to r = 6 m, since the point moves along a circle of radius 6 m. The speed of the point at time t = 1 s will be equal to v = 3 m/s, since v = 3t, and at t = 1 s, v = 3 m/s.

To find the acceleration, you need to use the formula for radial acceleration a = v^2/r. Substituting the known values, we get a = (3 m/s)^2/6 m = 1.5 m/s^2.

Now you need to find the angle between the acceleration and velocity vectors. To do this, you can use the formula cos(angle) = (av)/( |a||v| ), where |a| and |v| - modules of acceleration and velocity vectors, respectively.

Substituting the known values, we get cos(angle) = (1.5 m/s^2 * 3 m/s) / (1.5 m/s^2 * 3.16 m/s) ≈ 0.86. From the cosine table we find that the angle between the vectors is 26.6 degrees.

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