Solution to problem 7.8.7 from the collection of Kepe O.E.

7.8.7 Acceleration of point a = 1 m/s

Given: the acceleration vector of the point a = 1 m/s, the angle between the acceleration and velocity vectors is 45°, the radius of curvature of the trajectory r = 300 m.

Find: speed in km/h.

Answer:

Let us denote the speed of point a by v, and the angle between the acceleration and velocity vectors by α.

Then the projection of acceleration onto the velocity axis will be equal to:

a_v = a cosα = 1 * cos 45° = 0.707 м/с²

Taking into account that a_v = v²/r, find the speed v:

v = √(a_v * r) = √(0.707 * 300) ≈ 23.53 m/s ≈ 84.7 km/h

Answer: the speed of point a is about 84.7 km/h.

Solution to problem 7.8.7 from the collection of Kepe O..

This digital product is a solution to problem 7.8.7 from the collection of problems in physics by Kepe O.. The solution is made in accordance with the curriculum and contains a detailed description of the solution steps, as well as the answer to the problem.

Product characteristics:

• Title: Solution of problem 7.8.7 from the collection of Kepe O..
• Author: Kepe O..
• Product type: Digital product
• Russian language
• File format: HTML

By purchasing this product, you receive a ready-made solution to the problem, which can be used to prepare for exams or to improve your knowledge in the field of physics. The product is designed in a beautiful html format, which makes it easy to read and allows you to quickly find the necessary information.

This digital product is an indispensable assistant for everyone who studies physics and wants to successfully cope with tasks.

This product is a solution to problem 7.8.7 from the collection of problems in physics by Kepe O.?. The solution is made in accordance with the curriculum and contains a detailed description of the solution steps, as well as the answer to the problem.

To solve the problem, the known acceleration vector of point a, which is equal to 1 m/s, is used, as well as the angle between the acceleration and velocity vectors, which is 45°. The radius of curvature of the trajectory is also specified and is equal to 300 m.

Using the projection of acceleration onto the velocity axis, we obtain the value av, which is used to find the velocity of point a using the formula v = √(av * r). As a result, we obtain the speed of point a, which is about 52.4 km/h.

This digital product is an indispensable assistant for everyone who studies physics and wants to successfully cope with tasks. The product is designed in a beautiful html format, which makes it easy to read and allows you to quickly find the necessary information.

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Problem 7.8.7 from the collection of Kepe O.?. is to determine the speed of a point moving along a trajectory of a radius of curvature of 300 meters, provided that its acceleration is 1 meter per second and forms an angle of 45 degrees with the speed.

To solve this problem, it is necessary to use the formula for the radius of curvature of the trajectory:

R = (v^2) / a,

where R is the radius of curvature, v is the speed, and is the acceleration.

You also need to use the formula to convert meters per second to kilometers per hour:

v km/h = v m/s * 3.6.

Substituting known values, we get:

300 = (v^2) / 1,

v = √300 m/s.

Convert meters per second to kilometers per hour:

v km/h = √300 * 3.6 ≈ 52.4 km/h.

Thus, the speed of the point is approximately 52.4 km/h. The answer matches the answer given in the problem.

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