Solution to problem 7.7.17 from the collection of Kepe O.E.

7.7.17

For a given equation of motion of a point along a trajectory s = 0.6t², it is necessary to calculate the normal acceleration of the point at the time when the coordinate of the point is s = 30 m, and the radius of curvature of the trajectory is ? = 15 m. Round the answer to two decimal places.

Answer:

First, let’s find the derivative of the equation of motion of a point with respect to time:

v = ds/dt = 1,2t

Then we find the value of the speed of the point at the moment of time when its coordinate s is equal to 30 m:

v = sqrt(2as), where a is the normal acceleration of the point

30 = 0.6t²

t = sqrt(50)

v = 1.2sqrt(50)

Finally, we find the value of the normal acceleration of the point:

a = v²/? = (1,2sqrt(50))²/15 = 4.8 m/c²

Answer: 4.80

Solution to problem 7.7.17 from the collection of Kepe O..

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We present to your attention the solution to problem 7.7.17 from the collection of problems in physics by Kepe O.?.

First, you need to find the derivative of the equation of motion of a point with respect to time in order to find the value of the point’s speed:

v = ds/dt = 1,2t

Then we find the value of the speed of the point at the moment of time when its coordinate s is equal to 30 m:

v = sqrt(2as), where a is the normal acceleration of the point

30 = 0.6t^2

t = sqrt(50)

v = 1.2sqrt(50)

Finally, we find the value of the normal acceleration of the point:

a = v^2/?, where ? - radius of curvature of the trajectory

a = (1.2sqrt(50))^2/15 = 4.8 м/c^2

Answer: 4.80.

The electronic product contains a detailed description of the solution to the problem and a beautiful design in HTML format, which can be useful to students and teachers who study physics and want to improve their knowledge in this area. The product can be easily downloaded to your computer or mobile device and used at any convenient time.

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Solution to problem 7.7.17 from the collection of Kepe O.?. consists in determining the normal acceleration of a point along a trajectory specified by the equation of motion s = 0.6t^2, at the moment of time when its coordinate is s = 30 m and the radius of curvature of the trajectory is ? = 15 m.

To solve the problem, it is necessary to find the derivative of the point’s motion with respect to time, then calculate the value of the acceleration at time t, when s = 30 m. Next, it is necessary to find the value of the normal acceleration of the point using the formula:

a_n = v^2 / p

where v is the speed of the point at a given time, and p is the radius of curvature of the trajectory at this point.

Substituting the known values, we get:

s = 0,6t^2 v = ds/dt = 1,2t p = 15 м s = 30 м

From the equation of motion we find the moment of time t when s = 30 m:

30 = 0.6t^2 t^2 = 50 t ≈ 7.07 s

Let's find the speed of the point at this moment in time:

v = 1,2t ≈ 8,49 м/c

Now we can find the normal acceleration of the point:

a_n = v^2 / p ≈ 8,49^2 / 15 ≈ 4,8 м/c^2

Answer: 4.80.

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