If you look external the home window of a relocating train, you will certainly observe that an additional train lying stationary appears to be relocating in a backward direction. How does a stationary train seem to move? Behind it, tright here lies a very vital principle of family member activity, which will make us understand also why objects show up to relocate differently through different frames.

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What is Relative Motion?

The idea of referral frames was first introduced to talk about loved one activity in one or even more dimensions. When we say a things has actually a particular velocity, then this velocity is through respect to some structure that is known as the reference framework. In daily life, as soon as we measure the velocity of an object, the reference frame is taken to be the ground or the earth.

For instance, if you are travelling in a train and the train is moving at a rate of 100 km/hr, then your rate according to an additional passenger sitting on that train is zero. According to him, you are not moving. But if someone observes you from exterior the train, standing on the ground, according to him, you are relocating via 100 km/hr as you are on the train and the train is relocating via 100 km/hr.

Here, the movement oboffered by the observer relies on the location (frame) of the observer. This form of activity is dubbed loved one movement.

Relative Velocity

The relative velocity of an item A with respect to object B is the price of readjust of position of the object A with respect to object B.

If VA and also VB be the velocities of objects A and B through respect to the ground, then

The family member velocity of A with respect to B is VAB = VA – VB

The relative velocity of B through respect to A is VBA = VB – VA

Relative Motion in One Dimension

In one-dimensional motion, objects move in a straight line. So tright here are just two feasible cases:

Objects are relocating in the same directionObjects are relocating in the oppowebsite direction

Aget take the example of a man sitting on the train if the train is moving with 100 km/hr forward. Then according to the guy sitting on the train, the trees outside are moving backwards via 100 km/hr.

Since from the man’s point of see, the outside atmosphere is moving in the opposite direction to the train with the exact same velocity.

So for all forms of concerns, if you have to find the velocity of A with respect to B, then assume that B is at rest and also give the velocity of B to A in the opposite direction.

Relative Motion in Two Dimensions

The very same principle will be applicable in two-dimensional movement. If you have to discover the velocity of A through respect to B, assume that B is at remainder and also offer the velocity of B to A in the oppowebsite direction.

Let us take into consideration two objects A and also B which are relocating with velocities Va and also Vb via respect to some prevalent framework of reference, say, with respect to the ground or the earth. We have to uncover the velocity of A via respect to B, so assume that B is at remainder and also give the velocity of B to A in the opposite direction.

Vab = va – vb

Similarly, for the velocity of object B through respect to A, assume that A is at remainder and also provide the velocity of A to B in the opposite direction.

Vba = vb – va

Relative Motion Problems

1) Two bodies A and also B are travelling with the very same speed 100 km/hr in opposite directions. Find the family member velocity of body A with respect to body B and also loved one velocity of body B via respect to body A.

The loved one velocity of A w.r.t. B is VAB = VA – VB

= 100- (-100)

= 200 km/hr

Relative velocity of B w.r.t. A is VBA = VB – VA

= -100 – (100)

= -200 km/hr (-ve implies towards left)

In the very same question, if both bodies are relocating in the exact same direction via the very same speed then,

The relative velocity of A through respect to B is VAB = VA – VB

= 100-100

= 0

The relative velocity of B with respect to A is VBA = VB – VA

= 100-100

= 0

That means A is at rest with respect to B and also B is at remainder through respect to A, however both are relocating through 100 km/hr through respect to the ground.

2) Find the relative velocity of rain with respect to the moving man:

Here the guy is walking in the direction of the west with velocity Vm⃗vecV_mVm​​and also the rain is falling vertically downward with velocity Vr⃗vecV_rVr​​

So, the family member velocity of rain w.r.t. guy is Vm⃗=Vr⃗vecV_m=vecV_rVm​​=Vr​​

We recognize that the magnitude of the vector difference is provided by,

Vrm=Vr2+Vm2+2VrVmcos900=Vr2+Vm2Vrm =sqrtV_r^2+V_m^2+2V_rV_mcos90^0= sqrtV_r^2+V_m^2Vrm=Vr2​+Vm2​+2Vr​Vm​cos900

​=Vr2​+Vm2​​(from the diagram, Vm‾overlineVmVm is the hypotenusage of the triangle.

It θ is the angle which  Vrm‾overlineVrmVrm provides with the vertical direction then,

tan θ = BD/OB = Vm / Vr

In the above case, if the man desires to defend himself from the rain, he should host his umbrella in the direction of the family member velocity of rain through respect to the man. i.e.the umbrella must be organized making an angle θ ( θ=tan−1VmVr heta =tan^-1fracV_mV_rθ=tan−1Vr​Vm​​) west of vertical.

is the angle of the umbrella from the vertical.

3) Boat and also river difficulty. Find the velocity of the watercraft via respect to the river.

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Here the velocity of the watercraft through respect to the water or the velocity of the watercraft in still water is offered. If the observer is observing the activity from the ground, then the velocity of the boat through respect to the ground is equal to the velocity of the boat in still water plus the velocity of the water.

i.e Velocity of a boat through respect to the ground = velocity of the boat in still water + velocity of the water with respect to the ground