Let us consider a material point with mass m = 20 kg moving along a horizontal straight line. For this point there is a resistance force R, expressed by the formula:
R = 0,2v2,
where v is the speed of the point in m/s.
It is necessary to find the time during which the speed of the point decreases from 10 to 5 m/s.
To solve the problem, we use the equation of motion:
m(dv/dt) = -R,
where t is the time that has passed since the point began to move, and dv/dt is the rate of change of the point’s speed.
Substituting the expression for the resistance force, we get:
m(dv/dt) = -0,2v2.
Dividing both sides of the equation by m and integrating, we get:
∫(dv/v2) = -0,2/m ∫dt,
where ∫ is the sign of the integral.
Integrating, we get:
-1/v = 0,2/m t + C,
where C is the integration constant.
From the initial conditions of the problem it follows that at t = 0 the speed of the point is 10 m/s. Substituting these values, we find the value of the constant C:
-1/10 = 0,2/m * 0 + C,
C = -0,1.
Now you can find the time during which the speed of the point will decrease to 5 m/s:
-1/5 = 0,2/m t - 0,1,
where we get:
t = 10 seconds.
So, this problem is solved. We used the equation of motion and the laws of integration to express the time it would take for the point's speed to decrease from 10 m/s to 5 m/s.
We present to your attention a unique digital product - the solution to problem 13.2.23 from the collection of Kepe O.?. This product will be useful for anyone studying physics and mathematics, and in particular for those who are faced with problems involving acceleration and drag force.
In this product you will find a detailed solution to problem 13.2.23, which concerns a material point of mass m = 20 kg moving along a horizontal straight line. Based on the equation of motion and the laws of integration, we find the time during which the speed of the point will decrease from 10 to 5 m/s.
Our solution is presented in a beautiful html design, which makes it pleasant to read and easy to understand. You can use this solution to study a topic on your own or to prepare for an exam.
Purchase our digital product with the solution to problem 13.2.23 and expand your knowledge in the field of physics and mathematics!
We present to your attention a digital product - a solution to problem 13.2.23 from the collection of Kepe O.?.
To solve the problem, we considered a material point with a mass m = 20 kg, moving along a horizontal straight line under the action of a resistance force R, expressed by the formula R = 0.2v2, where v is the speed of the point in m/s. We needed to find the time during which the speed of the point will decrease from 10 to 5 m/s.
To solve the problem, we used the equation of motion: m(dv/dt) = -R, where t is the time that has passed since the point began to move, and dv/dt is the rate of change of the point’s speed. Substituting the expression for the resistance force, we obtained the equation: m(dv/dt) = -0.2v2.
Dividing both sides of the equation by m and integrating, we got: ∫(dv/v2) = -0.2/m ∫dt, where ∫ is the sign of the integral. Integrating, we got: -1/v = 0.2/m t + C, where C is the integration constant.
From the initial conditions of the problem it follows that at t = 0 the speed of the point is 10 m/s. Substituting these values, we found the value of the constant C: -1/10 = 0.2/m * 0 + C, C = -0.1.
Now we could find the time during which the speed of the point will decrease to 5 m/s: -1/5 = 0.2/m t - 0.1, from which we got t = 10 seconds.
In our solution to the problem, we used the equation of motion and the laws of integration to express the time during which the speed of a point will decrease from 10 to 5 m/s. The solution is presented in a beautiful html design, which makes it pleasant to read and easy to understand.
Purchase our digital product with the solution to problem 13.2.23 and expand your knowledge in the field of physics and mathematics!
We present to your attention a unique digital product - the solution to problem 13.2.23 from the collection of Kepe O.?.
This problem concerns a material point with mass m = 20 kg moving along a horizontal straight line. For this point there is a resistance force R, expressed by the formula: R = 0.2v^2, where v is the speed of the point in m/s. It is necessary to find the time during which the speed of the point decreases from 10 to 5 m/s.
To solve the problem, we used the equation of motion: m(dv/dt) = -R, where t is the time that has passed since the point began to move, and dv/dt is the rate of change of the point’s speed. Substituting the expression for the resistance force, we get: m(dv/dt) = -0.2v^2.
Dividing both sides of the equation by m and integrating, we get: ∫(dv/v^2) = -0.2/m ∫dt, where ∫ is the sign of the integral. Integrating, we obtain: -1/v = 0.2/m t + C, where C is the integration constant.
From the initial conditions of the problem it follows that at t = 0 the speed of the point is 10 m/s. Substituting these values, we find the value of the constant C: -1/10 = 0.2/m * 0 + C, C = -0.1.
Now you can find the time during which the speed of the point will decrease to 5 m/s: -1/5 = 0.2/m t - 0.1, from which we get: t = 10 seconds.
Our solution is presented in a beautiful HTML design, which makes it pleasant to read and easy to understand. You can use this solution to study a topic on your own or to prepare for an exam. Purchase our digital product with the solution to problem 13.2.23 and expand your knowledge in the field of physics and mathematics! Answer: 10 seconds.
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Solution to problem 13.2.23 from the collection of Kepe O.?. consists in determining the time during which the speed of a material point will decrease from 10 to 5 m/s under the action of a resistance force R = 0.2v2.
To solve this problem, it is necessary to use the equation of motion, which takes into account the resistance force:
m*a = F - R,
where m is the mass of a material point, a is its acceleration, F is the force acting on the point, R is the resistance force.
Since the material point moves along a horizontal straight line, then a = 0, therefore:
F = R.
Knowing the expression for the resistance force, we can write:
F = 0,2v^2,
where v is the speed of the material point.
Thus, we obtain the following equation:
m*dv/dt = 0,2v^2,
where dv/dt is the derivative of speed with respect to time.
Dividing both sides of the equation by v^2, we get:
(m/0,2)*dv/(v^2) = dt.
Integrating this equation from v1 = 10 m/s to v2 = 5 m/s, we obtain:
(m/0,2)*(-1/v2 + 1/v1) = t.
Substituting the numerical values m, v1 and v2, we get:
(20/0,2)*(-1/5 + 1/10) = t,
whence t = 10 seconds.
Thus, the speed of the material point will decrease from 10 to 5 m/s in 10 seconds under the action of a resistance force R = 0.2v^2.
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IDZ 19.1 – Option 9. Solutions Ryabushko A.P. - В результате проведенного эксперимента были получены… ...