Mechanics Simple Guidance For In Physics Motions Graphs.

Kinematics

Kinematics is a branch of classical mechanics that describes the motion of points, bodies or any other objects, without considering the mass of each of the forces that caused the motion.

There are two distinct types of motion.

  1. One-dimensional motion.
  2.  Two-dimensional motion.

One-dimensional motion.

It involves anything that moves along a straight line. Velocity, acceleration, and time are defined before. They used in four equations for motion like this. We know already.v=u+at,\,\,v^2=u^2+2as,\,\,s=ut+\frac{1}{2}at^2,\,\,s=(\frac{u+v}{2})t

Here the acceleration is constant in all the problems. Free-falling bodies are examined as an example of one-dimensional motion.

Two-dimensional motion.

Here, as an example, we can consider the projectile motion of a body we use the above equations, with the addition of vectors and components. Let’s we learn more about this later. Now we continue with relative motion. 

Certain frame

In this lesson, A certain frame can be defined in the form of either a basic procedure or a frame structure. It can be considered as the frame where the observer is taking the reading for the experiment. Think for a moment,
I am the observer. When I’m standing on the road we can see vehicles are moving. I’m standing on the road, but I’m(c) not moving as shown in the fig. When I in the place of “A” the certain frame will be the A and its surroundings. At that time, I am the observer. When I in the place of “B” the certain frame will be the B and its surroundings.

Relative motion

Ok, Now you know what is the certain frame. Now we going to talk about relative motion.

Relative motion is the calculation of the motion of an object with regard to some other moving object. According to the calculations of relative motions, we can be divided them into the three fields such ass following.

  1. Motion in the same direction.
  2. Motion in the opposite directions.
  3. Motion when a θ angle is formed between directions of A and B.

Motion in the same direction

We consider two objects A and B moving in the same direction in parallel paths. The velocities of A and B are respectively VA and VB. lets we obtain the velocity of A at the sight of an observer at B. Here,

The velocity of A relative to B = (x-y)  its direction is to the → left to right as shown in the fig. If x>y the direction of (x-y) is the same → left to right. If x<y the direction of (x-y) is the same ← left to right because the (x-y) <0. If (x-y) =0 Observer can see the A is not moving. Now we understand the resulting velocity of A relative to B is (x-y) when A and B are same directions.

Motion in the opposite directions.

A and B moving in the opposite direction in parallel paths. The velocities of A and B are respectively VA and VB same as the above. Lets we obtain the velocity of A at the sight of an observer at B. Here, the velocity of A relative to B is always (x+y) because of the position of the opposite. There is no zero value for the (x+y). You want to understand how is the equation shown in the fig. AB=AE+EB. We can replace E with any certain frame.

Example1

  1. A boat has a speed of 3.0ms-1 in still water. It is used to cross a river 1000m wide along. The water flows speed at 1.8ms-1 in the river. The boat is directed towards the opposite bank but the water carries it downstream. Calculate
    1.  the speed of boat relative to the bank of the river,
    2. the distance the boat is carried downstream from its point of departure to its point of arrival at the opposite bank,
    3. the time is taken to cross the river.

Answer1

After reading the problem we drow the sketch for it as shown in this fig. Let we assume
B = Boat, W = Water and E = River Bank is meaning Earth. In the question, you can see
“A boat has a speed of 3.0ms-1 in still water”.
It’s mean BW = ↓3.0ms-1 . “The water flows speed at 1.8ms-1 ” It’s mean WE =↓1.8ms-1 . That’s only we can drag from the problem which helps us to solve it.

A. In the part A we want to find BE. Next, Draw the velocity triangle of BE = BW + WE

BE = \sqrt{3^2+(1.8)^2}=3.49\,ms^{-1}

direction = angle between BE and the bank
angle=tan^{-1}(\frac{3}{1.8})=tan^{-1}(\frac{5}{3})

B.\, The\, time\, for\, reach\, the\,point=\frac{1000}{3}
therefore \,the \,distance\, of\, downstreem=\frac{1000}{3}\times1.8=600m

C. \, \frac{1000}{3}=333.33s

Motion when a θ angle is formed between directions of A and B.

Now lets we learn how is this position. There many steps to solve any relative motion problems. So we want to follow them. Following are the steps.

  1. First, we want to understand the problem. To do that we should read carefully without any doubt.
  2. Scetch down the problem with graphically. It should be understandable. It should help you to solve the problem correctly. It should be contained with all data of the problem.
  3. Next we drow the velocity triangle for the problem.
  4. Now we get the equation which we want to solve it.
  5. Applying the values to the equation and get the answer.

We can solve any problem of relative motions by the above 5 steps.

For A, B, C’s three Certain frames,
(AB) = (AC)  + (CB) .
The speed of A and B are respectively
VAE = y and VBE = x.
The angle of y and x is θ. Now we want to find the velocity of A relative to B.
It is very easy,
you can see it is done in this fig.

Example2

  • Two ships, A and B, leave a port at the same time. Ship A travels due north at a steady speed of 10kmh-1 and ship B travels N600E at a steady speed of 15kmh-1 . 
    1. What is the distance and direction from A to B after 1 hour?
    2. What is the velocity of B relative to A?
    3. What is the direction of

Answer2

First, It is necessary to know about trigonometry rules for quick the solution. This is cosine rule of a trigonometry. In this fig, there is a rhythm between a, b, c, A, B, and C. If we understood it we can remember it well. By this, you can save the time.

 

  1. After a one hour, ship A has moved 10km to the north and ship B has moved 15km to the N600E. N600E mean, 60 is measured from North to East.
    distance and direction from A to B after 1 hour = AB (as shown in fig)
    Now use above formula,
    AB=\sqrt{10^2+15^2-2\times10\times15\times{cos60^0}}
    =13.2287km
  2. To find the velocity of B relative to A we want to
    draw the velocity triangle. In this fig, now we can say what are the values of the above formula.
    AB=\sqrt{10^2+15^2-2\times10\times15\times{cos60^0}}
    =13.2287kmh^{-1}
  3. Next, we can find the value of α
    α = angle between the North and velocity of B relative to A.
            tan{\alpha}=\frac{15sin{60^0}}{10-15cos60^0} =5.1961
    \alpha=\tan^{-1}{(5.1961)}
    \alpha=1.38radians
    \alpha=79degrees,\, 6minutes

The rectilinear motion under constant acceleration

 We consider the s-t graph of an object that moves in a straight line. In the s-t graph, velocity = slope of the line at a given point on the s-t graph. OA = the object moves at a steady speed because its slope is constant. The slope of a straight line is constant. It does not change with the time. There are so many examples. One example is a car moves at a steady speed on the straight road. 

In OCB graph, Its velocity is changing always with the time because the slope of the graph is changing with time always. The velocity at the time t1 = slope or gradient at the point C on the graph.

How can we find the gradient at the point on the graph

First, we should choose a point on the graph. Next, we plot an antenna on the point which graph and line t1C are intersected. Select two points as far away as possible on the antenna. Now we calculate the slope by using this formula. More information about graph

The\,gradient\,of\,the\,line\,= \,\frac {y_b-y_a}{x_b-x_a}

The gradient of the s-t graph = velocity

Velocity – time graph

The velocity at the time t1 = slope or gradient at the point F on the graph as shown in this fig. In this fig, you can see 3 types of v-t graphs.
They are DE, OA, and OFCB.

OA graph

OA line has a constant gradient. Therefore it moves constant acceleration. The acceleration = gradient of the v-t graph. A body starts from rest and with a uniform acceleration. Its velocity is continuously increasing with time. There are so many examples. One is a free falling body under the gravity.

DE graph

It has constant velocity. Or body is moving at a steady speed. Every time its acceleration is zero.

OFCB graph

Its gradient or slope changes with the time. This body is started with rest because the graph is started from the point of (0,0). Its mean velocity = 0, time = 0.  While we are seeing from point O to F on blue colored line, the slope of the graph is decreased from a positive value to the minimum value. You can see it well. Its mean the body is decreased its velocity from  V0  to  V within time t. Here,  t1  is in seconds.  V0  > V . From F to C, the slope of the graph is increased.  Its mean the velocity of the body is increased from V1  to  V2 within (t– t) seconds. Let’s we assume the velocities of the body at the points O, F, and C respectively   V0  , V, and  V2 .  Acceleration of this motion is changed with the time. It is positive from O to B. After B the gradient decreases from a positive value to the zero when the slope exactly parallel to the t- axis. After that gradient decreases from zero to negative.

The distance body has moved within the time of  (t– t) seconds = Average speed * Time

Distance\, =\frac{1}{2}(GF+CH)(t_2-t_1) =\frac{1}{2}(GF+CH)(GH)=area\, (FGHC)

The area between a velocity graph and t-axis represent the displacement of the body during the time.

A body starts with an initial velocity of u and is moving in a straight line with a constant acceleration of a. After a time of t, let’s we assume its velocity is v and its distance from the start is s. We can show this case in a v-t graph.
We can find the acceleration by the gradient of the graph. The gradient of the graph is tanθ. lets we find it,
tan\theta=a=\frac{v-u}{t}
then, \, v=u+at, The distance = area between the graph and t-axis
s=\frac{1}{2}(u+v)t
we can get the other equations also.
v=u+at,\,\,v^2=u^2+2as,\,\,s=ut+\frac{1}{2}at^2,\,\,s=(\frac{u+v}{2})t

An Example of Motions in a straight line.

  • A train travels in a straight line at a constant acceleration of 2 ms-2  .  The train is increased its speed from 10 ms-1 to v ms-1 within 30 s.
  1. Calculate the by using a v-t graph of this motion.
  2. Draw the s-t graph and a-t graph.

Answer:

1.
acceleration = gradient of the v-t graph
2=\frac{v-10}{30}
v=70ms^{-1}

 

2.
v-t graph = gradient of the s-t graph. Therefore we can draw the s-t graph according to its v-t.

We can find the distance also between the velocities of 10 and 70.

distance=\frac{1}{2}(30)(70-10)=900m

It has traveled 900m in 30 seconds.

Drawing the s-t graph:

First, scetch down the same two axes below the v-t graph as shown in this fig. Rename it’s two axes to the s – t. Now we find the value of the Point A on the graph s-t according to the Point A on the graph v-t. The v-axis value of point A on the graph v-t (10)=The gradient of the s-t graph at the point A=tan\theta_1.

The v-axis value of point B on the graph v-t (70)=The gradient of the s-t graph at the point B=tan\theta_2. tan\theta_1<tan\theta_2. Therefore we can decide how the curve should be plotted from A to B. It should increase its gradient or angle from A to B. You can see the orange colored line in the fig.

Drawing the a-t graph:

Now we want to draw the a-t graph from the v-t graph. Same as the above first we want to scetch down the two axes and name its axis as a and t as shown in the fig. Now we want to find the initial value, end value and how is the gradient or angle of the s-t graph between t=0 and t=30. We know already that, the gradient of the v-t graph = a-t graph.

In this case, we can see the gradient(2) of the v-t graph is constant and positive at all the time. It is simply. You can see the orange colored straight line in the fig.

More examples of linear motions.

  1. Choose the right expressions through the following expressions.
    1. An Object which travels with constant acceleration can have  a zero velocity
    2. The velocity of an object which moves at a steady speed can be changed with the time.
    3. The speed of an object which travels at a constant velocity can be changed with the time.
    4. An object accelerates to the West, has a velocity to the East.
    5. The direction of the object which travels with a constant acceleration can be changed.
  2. A motor bicycle which travels at a speed of 20 ms-1 is braked and stopped under the deacceleration of 1 ms-2 . Find the distance and the time which travels with deacceleration.
  3. A train which starts from rest at a station is traveled it’s first 2km at a constant acceleration and next, it is traveled 6km at a steady speed. Finally, it is traveled 1km with constant deacceleration until it comes to the rest. If the distance between the two stations is 12km and the time it has taken is 20minutes find the
    1. acceleration of the train by using v-t graph.
    2. steady speed.
    3. deacceleration
  4. Find the maximum height and flying-time of the throw of a small stone which is thrown vertically upward at an initial speed of 40 ms-1. Consider g = 10 ms-2 .
  5. A coconut falls from a tree to the ground within 4s. Find
    1. the height of the tree.
    2. the speed of the coconut just before impact with the ground.
  6. Find the flying time of a stone which is thrown vertically upward at an initial speed of 25 ms-1 from a top of a 50m high tower.
  7. The s-t graph of the object which travels to the x-direction is showing in the following fig. The acceleration or de-acceleration of the object is constant. Draw
    1. v-t graph
    2. a-t graph for this motion.
  8. Given below a-t graph indicates the way of changing the acceleration(a) of an object with the time(t). Drow a
    1. v-t graph
    2. s-t graph for this motion.
  9. Find the resultant velocity of an object which 10 ms-1 and 12 ms-1 are acted perpendicularly each other on the object.
  10. Two velocities of 8 ms-1 and 6 ms-1 are acted on an object with an angle of 450. Find the resultant velocity of the object.