The **linear velocity** It is defined as that which is always tangential to the trajectory followed by the particle, regardless of its shape. If the particle always moves in a rectilinear trajectory, there is no problem in imagining how the velocity vector follows this straight line.

However, in general the movement is carried out on a curve of arbitrary shape. Each portion of the curve can be modeled as if it were part of a circle of radius *to*which at every point is tangent to the trajectory followed.

In this case, the linear speed follows the curve tangentially and at all times at each point on it.

Mathematically, the instantaneous linear velocity is the derivative of the position with respect to time. Be ** r** the position vector of the particle at an instant

*you*then the linear velocity is given by the expression:

*v** = r‘*

*This means that the linear speed or tangential speed, as it is also usually called, is nothing more than the change of position with respect to time.*

*[toc]*

**Linear velocity in circular motion**

**Linear velocity in circular motion**

*When the movement is on a circle, we can go next to the particle at each point and see what happens in two very special directions: one of them is the one that always points towards the center. This is the address radial.*

*The other important direction is the one that runs on the circumference, this is the direction tangential and the linear velocity always has it.*

*In the case of uniform circular motion, it is important to realize that the velocity is not constant, since the vector changes its direction as the particle rotates, but its magnitude (the size of the vector), which is the speed, yes it remains unchanged.*

*For this motion the position as a function of time is given by s*

*If the magnitude of the speed also varies (we already know that the direction always does, otherwise the mobile could not turn), we are facing a varied circular movement, during which the mobile, in addition to turning, can slow down or accelerate.*

**Linear velocity, angular velocity, and centripetal acceleration**

**Linear velocity, angular velocity, and centripetal acceleration***The motion of the particle can also be seen from the point of view of the swept angle, instead of doing it from the arc traveled. In such a case, one speaks of the angular velocity. For a movement on a circle of radius R.there is a relationship between the arc (in radians) and the angle:*

*s = Rθ*

*Differentiating with respect to time on both sides:*

*ds/dt = R(dθ**/dt)*

*Calling the derivative of θ with respect to you as angular velocity and denoting it with the Greek letter ω «omega», we have this relationship:*

*v = ω**R.*

**Centripetal acceleration**

**Centripetal acceleration***Every circular motion has centripetal acceleration, which is always directed towards the center of the circle. She ensures that the speed changes to move with the particle as it rotates.*

*centripetal acceleration ac either aR it always points to the center (see figure 2) and is related to linear velocity in this way:*

*ac = v2 /R*

*And with the angular velocity as:*

*ac = (ω**R )2 /R = ω**2R*

*For a uniform circular motion, the position s*

*s*

*Furthermore, varied circular motion must have a component of acceleration called tangential acceleration at, which deals with changing the magnitude of linear velocity. Yeah at it’s constant, the position is:*

*s*

*With you as the initial velocity.*

**Solved problems of linear speed**

*The solved exercises contribute to clarify the proper use of the concepts and equations given above.*

**-Exercise solved 1 **

**-Exercise solved 1***An insect moves on a semicircle of radius R = 2 m, starting from rest at point A while increasing its linear speed at a rate of pm/s2. Find: a) After how much time does it reach point B, b) The linear velocity vector at that instant, c) The acceleration vector at that instant.*

**Solution**

**Solution***a) The statement indicates that the tangential acceleration is constant and is worth π m/s2, so it is valid to use the equation for uniformly varied motion:*

*s*

*With so = 0 and vo = 0:*

*s*

* **s = π**R. *(Half the length of the circumference)

* **t = (2. π**R.** /**aT) ½ s = (2π**.2 /π**)**½**s = 2 s*

*b) v*

*When at point B, the linear velocity vector points in the vertical direction downward in the direction (- and):*

*v*

*c) We already have the tangential acceleration, we need the centripetal acceleration to have the velocity vector to:*

*ac =v2 / R = (**2π**)2 / 2 m/s2 =2π**2 **m/s2*

*to** = ac(-**x**) + aT (-**and**) = 2π**2**(-**x**)+ π** (-**and**) m/s2*

**-Exercise solved 2**

**-Exercise solved 2***A particle rotates in a circle of radius 2.90 m. At a particular instant its acceleration is worth 1.05 m/s2 in a direction such that it forms 32º with its direction of movement. Find its linear velocity at: a) This moment, b) 2 seconds later, assuming that the tangential acceleration is constant.*

**Solution**

**Solution***a) The direction of movement is precisely the tangential direction:*

*aT = 1.05 m/s2 . cos 32º = 0.89 m/s2 ; BC = 1.05 m/s2 . sin 32º = 0.56 m/s2*

*The velocity clears of ac = v2 / R as:*

* **v = (R.ac)1/2* *= 1.27m/s*

*b) The following equation for uniformly varied motion is valid: v = vo + aTt = 1.27 + 0.89 .22 m/s = 4.83 m/s*

**References**

**References***Bauer, W. 2011. Physics for Engineering and Science. Volume 1. Mc Graw Hill. 84-88. Figueroa, D. Physics Series for Science and Engineering. Volume 3rd. Edition. Kinematics. 199-232. Giancoli, D. 2006. Physics: Principles with Applications. 6th. Ed Prentice Hall. 62-64. Relative Motion. Recovered from: courses.lumenlearning.com Wilson, J. 2011. Physics 10. Pearson Education. 166-168.*