GLB dan GLBB

Mekanika-3 (glb dan glbb)

Konsep yang kita pelajari kali ini adalah tentang gerak lurus. Hampir semua benda yang kita jumpai mobil, kereta, motor dll, adalah contoh fenomena gerak lurus. Benda yang bergerak lurus ada yang memiliki kecepatan tetap (konstan) ada juga yang kecepatannya dipercepat dengan percepatan tetap (konstan).

Gerak benda dengan kecepatan tetap disebut GLB dan gerak benda dengan kecepatan yang berubah dengan teratur disebut GLBB.

1. GLB (gerak lurus beraturan)

          Ciri gerak ini v = tetap, artinya perubahan kecepatan = 0, sehingga a (percepatan) = 0. Artinya dalam selang waktu yang sama jarak yang ditempuh benda juga sama, persamaan yang digunakan dalam gerak ini pun sangat sederhana 

          S = V x t

          S = jarak (m)

          V = kecepatan benda (m/s)

          t = waktu (s)

          Berikut ciri grafik gerak GLB

1. GLBB (gerak lurus berubah beraturan)

          Ciri dari gerak lurus ini a = tetap, sehingga terjadi perubahan kecepatan yang teratur sesuai dengan percepatan benda. Percepatan yang dialami benda bisa bertanda positif (dipercepat) ataupun negatif (diperlambat). Persamaan yang digunakan cukup mudah diingat

          V = Vo + at

          S = Vo t + 1/2 at² 

          V² = Vo² + 2as

 

yang perlu diingat adalah " tanda percepatannya apakah dipercepat (+) atau diperlambat (-) " 

         Berikut ciri grafik dari gerak GLBB

 

Contoh Soal :

1. Dari gambar dibawah tentukan :

 

    a. jarak yang ditempuh benda dari t = 20 s sampai t = 60 s

    b. percepatan benda dari t = 0 s sampai t = 20 s

    c. perlambatan benda dari t = 60 s sampai t = 80 s

    d. jarak total benda dari mulai bergerak sampai berhenti

Penyelesaian : 

a. Trick : " jarak (S) = luas daerah persegi panjang "

     S = 40 x 15

        = 600 m

  

b. Trick : " a = kemiringan grafik (nilai tan α) "

     a =  (15 - 0) / (20 - 0)

        =  0,75 m/s² , (+) ---> dipercepat

c. sama dengan trick sebelumnya

    a = (0 - 15) / (80 - 60)

       = - 15 / 20

       = - 0,75 m/s² , (-) ---> diperlambat

d. Trick : "  S = luas daerah Trapesium "

      S = 1/2 x (40 + 80) 15

         =  900 m

2.  Sebuah sedan berada di depan sebuah bus dengan jarak 50 m, keduanya sedang melaju di jalan tol dengan kecepatan yang sama. Suatu saat sedan diperlambat dengan perlambatan 3 m/s² sedangkan bus yang berada di belakangnya diperlambat dengan perlambatan 2 m/s² karena di depan kedua kendaraan telah terjadi kecelakaan. Berapakah kecepatan bus sesaat sebelum menabrak sedan ?

Penyelesaian :

Trick : “ jarak yang ditempuh bus (Sb) = 50 m + jarak yang ditempuh sedan sampai berhenti (Ss) “

Sb = 50 + Ss

Vo + ½ a_b t² = 50 + Vo + ½ a_s t²

- ½ 2 t² = 50 - ½ 3t²

½ t² = 50

t = 10 s

sedan

0 = Vo – 3 . 10

Vo = 30 m/s (Vo untuk sedan dan bus adalah sama)

Bus

V = 30 – 2 . 10

V = 10 m/s (kecepatan bus sesaat sebelum menabrak sedan)

picture;

 

Downlad soal glb dan glbb

Download Ppt File

Problems for Motion with Constant Acceleration in Two and Three Dimensions

Problem : A basketball player who is standing 15 feet away from a basketball hoop is trying to make a basket. If the height of the hoop is 10 feet, and the height at which the player shoots the ball is 6 feet, at what angle and with what speed should the player shoot the ball?

Figure %: Diagram of a basketball player shooting a ball with speed v and angle θ .

Solution for Problem 1 >>

Using the general formula for projectile motion worked out in this section, we find that the position function for a ball which is shot from a height of 6ft, with initial velocity vector v ₀ , is:

x(t) = (0, 0, - g)t ² + v ₀ t + (0, 0, 6)

where we are taking the origin of coordinates to be at the player's feet, and the positive z -direction points upwards. If we choose the x -direction to point in the direction of the hoop, we can ignore the y -direction and reduce this to a two-dimensional problem.

For the player to make his basket, the trajectory of the ball must pass through the point (15, 0, 10) (since the basket is located 15 ft away and is 10 ft high). Therefore, we need to find v ₀ = (v _x, 0, v _z) such that x(t) = (15, 0, 10) at some time t (since there is no motion in the y -direction we know that v _y must be zero). To do this, we write out the one-dimensional equations for the x - and z-directions:

(1) 15 = v _x t , and (2) 10 = -  gt ² + v _z t + 6

From (2) we find that the ball reaches a height of 10 ft at a time t such that:

 gt ² - v _z t + 4 = 0

Using the quadratic formula, we find two solutions for t :

t ₁ =  and t ₂ = .

t ₂ , which is the earlier time, corresponds to when the ball passes through a height of 10 feet on its way up. However, the basket must be made as the ball is falling back down. Therefore, t ₁ is the time we are interested in.

It is not enough for the ball to have a height of 10 ft, however, in order for the basketball player to make the basket. At the same time it reaches that height (i.e. at time t ₁ ) it must also have traveled 15 ft in the x - direction. We are now in a position to solve for v _x in terms of v _z . Using (1), and plugging in the timet ₁ , we find that:

v _x =  = 

So, any initial velocity vector of the form

v ₀ = (, 0, v _z)

will ensure that the player makes his basket.

Figure %: Several solutions to the problem of a basketball player shooting to make a basket.

In other words, for every possible value of v _z (i.e. for every value of v _z such that  is a real number), there is a solution. You were probably already aware of this intuitively because by shooting a basketball at different angles and with different speeds you are able to make the same basket in a variety of ways!

Close

Problem : Assume that the basketball player in the previous problem has a special condition which allows him only to shoot basketballs at a speed of 10 ft/s. How might someone go about finding at what angle must he shoot the ball in order to make the basket?

Using the solution to the previous problem,

v ₀ = (, 0, v _z)

we need only require that the magnitude of v ₀ is 10 ft/s. This will enable us to solve for a single value of v _z , which will in turn provide us with a unique value for v _x . Following familiar rules from trigonometry, by taking arctanv _z/v _x we can then get the angle at which the player must shoot. The details of these particular calculations become quite ugly and are left as an exercise.

The Uniform Acceleration Motion

his is a motion during which the instantaneous acceleration is always constant.

(1)

A constant acceleration reveals that:
1)the direction of motion is along a straight line.
2)Average acceleration and instantaneous acceleration have the same magnitude:

By coupling the last relationship with the definition of average acceleration, we can build the velocity law for a uniform acceleration motion



If  then , and the last one can be written as

(2)

The velocity versus time graph is a straight line and its angular coefficient is the motion’s acceleration .

From this graph the relationship between the average velocity and the instantaneous velocity can be deduced. It follows from the idea of a segment’s average point:

By coupling this with the previous equation (2) and with the definition of average speed, we have a three equation system. Its solution represents the law of motion for uniform acceleration motion:

By inserting the (b) into the (a) we have:

and by inserting the (c) into the last one:

that leads to:

(3)

The graph of this motion law is a branch of a parabola. For instance, if the initial distance  and the initial velocity  are equal to zero, the diagram is similar to the following one:

The parabolic behaviour is strictly related to the square dependence on time

The Uniform Velocity Motion.

This is a motion during which the instantaneous velocity is always constant.
Hence acceleration is zero.

	(1)
	(2)

A constant velocity reveals that:
1)the direction of motion is along a straight line.
2)Average velocity and instantaneous velocity have the same magnitude

By coupling the last relationship with the definition of average velocity, we can build the motion law for all the uniform velocity motions.



If  then , and the last relationship can be written as

(3)

As this law shows, the distance versus time graph is a straight line and its angular coefficient is the speed of motion 
.

Uniformly Accelerated Motion (Graphical Treatment)

he velocity-time graph for uniformly accelerated rectilinear motion (motion along a straight line), a > 0, is illustrated in the figure.

The velocity-time graph for a uniformly retarded motion is as shown in the figure. Here a < 0, indicating retardation or deceleration.

In this section, we are going to derive the velocity-time relation for a particle moving with constant acceleration.

The figure above illustrates a particle moving with constant acceleration 'a' along the positive direction of X-axis. If v (t₁) and v (t₂) be the velocities of the particle at times t₁ and t₂ respectively, then

Acceleration, according to the definition

Cross-multiplying, we have,

v (t₂) - v (t₁) = a (t₂ - t₁)

 v (t₂) = v (t₁) + a (t₂ - t₁)

If the motion is considered from t = 0, then the corresponding velocity v(0) at t = 0 is called the initial velocity and the velocity after a time interval, say t, is called the final velocity and denoted by v(t). In this case, the acceleration is given by

or at = v (t) - v (0)

v(t) = v (0) + at

The above expression represents the first equation of motion.

The figure below illustrates the velocity-time graph AB of a uniformly accelerated particle. The points A and B correspond to velocities v (0) and v (t) respectively.

Slope of the straight line AB

Also, from the definition of acceleration, we know that

 We conclude that the slope of the velocity-time graph for uniformly accelerated motion gives the acceleration.At any given time, the direction of motion is given by velocity and not by acceleration. As an example, when a ball is thrown up vertically, its velocity is directed upwards at any given time, but its acceleration is always directed in the downward direction.