Motion In a Straight Line - 11th physics

Velocity-Time Graph for Some Simple Cases

Case (a) Motion of an object with positive direction and positive acceleration.

Case (b) Motion of an object with positive direction and negative acceleration.

Case (c) Motion of an object with negative direction and negative acceleration.

Case (d) Motion of an object with negative acceleration that changes its direction at time t₁. Between times 0 and t₁, it moves in positive x–dirction and between time t₁ and t₂ it moves in the opposite direction.

Calculation of Displacement using velocity-time graph

In velocity-time graph of a moving object the area under the curve represents the displacement over a given time interval.

Let a velocity-time graph of an object which is moving with a constant speed u.

Again, let this object is moving between time t = 0 and t = T

Thus, Velocity-Time graph of this moving object is

Thus, distance covered by this moving object = Area under the rectangle formed in the graph of height u and base T

= Length × Height

= u × T = uT

Thus, Area of the rectangle = uT

This Area of the rectangle = Distance covered by the moving object = uT

This area uT is equal to the displacement in the given time interval of 0 and T.

Thus, using velocity-time graph the displacement of the moving object can be calculated.

It should be noted that physically acceleration and velocity cannot change values abruptly at an instant. Rather Changes are always continuous.

Kinematic Equations For Uniformly Accelerated Motion

Let a particle is moving with constant acceleration a.

Again let the initial velocity at time 0 is equal to u

And the velocity at time t = v

Therefore, Acceleration

`a=(dv)/(dt)`

`=>dv =a\ dt `

After integrating both sides, we get

` int_(u)^(v) dv = int_(0)^(t)adt`

Here as time changes from 0 to t the velocity changes from u to v.

Thus after evaluating the integral, we get

` [v]_u^v = a[t]_0^t `

⇒ v – = at

⇒ v = u + at - - - - - (i)

Now, above equation can be written as

` (dx)/(dt) = u + at`

`=> dx = (u +at)dt`

After integrating both side

` => int_0^xdx=int_0^t(u+at)dt`

[At t = 0 the particle is at x = 0. As time changes from 0 to t the position changes from 0 to x.]

` =>[x]_0^x=int_0^tu\ dt + int_0^tatdt `

`=>x=uint_0^t\dt+a\int_0^t\tdt`

`=>x=u[t]_0^t+a[t^2/2]_0^t`

`=>x=ut+1/2at^2` - - - - - (ii)

Now, from equation (i) we have

v = u + at

After squaring both sides, we get

v² = (u + at)²

⇒ v² = u² + 2 uat + a²t²

⇒ v² = 2a(ut + 1/2 at²)

[After substituting the value x= ut + 1/2 at² from equation (ii)]

⇒ v² = u² + 2ax - - - - (iii)

Thus, we get three equations for an object in motion in a straight line with constant or uniform acceleration.

`v=u+at`

`x=ut+1/2at^2`

`v^2=u^2+2ax`

The values of quantities initial velocity (u), final velocity (v) and acceleration (a) are taken positive or negative depending on the direction along the positive or negative.

Free Fall

If an object is released under the earth gravity from a height and air resistance is considered neglected, the object falls downwards, this is called Free Fall.

In the free fall the magnitude of acceleration of the falling object is represented by g. And g, the gravity of earth is taken to be equal to 9.8 ms^–2.

Thus free fall is a case of motion in a straight line with constant acceleration.

In the case of downwards fall, the gravity (g) is taken negative and is equal to – 9.8 ms^–2

But when an object is thrown straight in upward direction, the gravity of earth (g) is taken as positive.

Relative Velocity

Velocity of one object in respect with other one is called relative velocity.

Let two object A and B are moving uniformly with average velocities v_A and v_B in one dimension, say along x-axis in the same direction.

And if x_A(0) and x_B(o) are positions of objects A and B, respectively at time t=0, thus, their position x_A(t) and x_B(t) at time time t are given by:

x_A(t) = x_A(o) + v_At

x_B(t) = x_B(o) + v_Bt

Then, the displacement from object A to object B is given by

x_BA(t) = x _B(t) – x _A(t)

= [x_B(o) – x _A(o)] + (v_B – v_A)t

Above equation tells that as seen from object A, object B has a velocity v_B – v_A because the displacement from A to B changes steadily by the amount v_B–v_A in each unit of time.

Thus, we say that the velocity of object B relative to object A is v_B–v_A

That is v_BA=v_B – v_A

Similarly, velocity of object A relative to object B is

v_AB=v_A – v_B

This shows that, v_BA = –v_AB

Three Cases of Relative Velocity

Let two objects A and B are in motion in same direction along a straight line and their velocities are v_A and v_B.

Case–I: When Velocity of two Objects are Same

Now, if v_A (Velocity of object A) = v_B (velocity of object B)

Therefore the displacement of object (x_BA)

= x_B(t)–x_A = x_B(0) – x_A(0)

This means that the displacements of objects B and object A after time t are equal.

That is the given two objects stay at a constant distance (x_B – x_A(0)) apart.

In this case the velocities of both of the objects are equal, the position–time graphs are straight line parallel to each other. And the relative velocity v_BA or v_AB is zero in this case.

Case–II: When Velocity of Object B is greater than the Velocity of Object A

In this case the velocity of object B is greater than the velocity of object A

This means, v_B > v_A

This, v_B–v_A is negative.

In this case in the velocity-time graph the graph of one object is steeper than the other and they meet at a common point.

Case–III: When Velocities of Objects have opposite sign

If velocities of objects have opposite signs. This means that the velocity of one object is positive while the velocity of other object is negative.

This means, the velocity of object B = –v_B

And the velocity of object A = v_A

This means objects are moving in opposite direction.

In this case of velocities having opposite signs in position-time graph, the both graphs meet at a common point.

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