• Motion and rest are relative terms.
• For example, suppose a box kept inside a moving train is observed by a person standing on the ground. For that person, the box is moving with the speed of train.
• The same box is observed to be at rest by a passenger traveling in the train. 

• It concludes that for an observer, he himself is not in motion even if he is moving with respect to someone else.
• This concept is very useful in understanding and analyzing the motion of different bodies.
• As one object is at rest thus, number of equations can be reduced. 

• Consider a situation in which a passenger is moving in straight line with constant velocity \(\text v_0\hat {\text i}\) on the platform.
• A train is also moving with constant velocity  \(\text v_{\text t} \hat {\text i}\), as shown in figure. 
• The velocities shown are in reference frame attached to the ground.
• Now, the reference frame is attached to the passenger. He observes that the velocity of train is not same as before.
• Because, according to passenger, his velocity is zero.
• To make him at rest in frame of reference attached to him, add  \(-\vec{\text v}_0\)  to all objects.

\(\vec {\text v}_0=\text v_0\hat {\text i}=\) velocity of passenger in ground frame

\(\vec {\text v}_{\text t}= {\text v}_{\text t}\hat {\text i}=\) velocity of train in ground frame

\(\vec {\text v}_{{\text t}/0}=\) velocity of train observed by passenger

\(\vec {\text v}_{{\text t}/0}=\vec{\text v}_{\text t}-\vec {\text v}_0\)

\(\vec {\text v}_{{\text t}/0}={\text v}_{\text t}\hat {\text i}- {\text v}_0\hat {\text i}\)

Relative Motion in One-Dimension

• Two men \(\text m_1\) and \(\text m_2\) are moving along the same line, as shown in figure.
• They both have different velocities, \(\vec {\text v}_1=3 \,\text {m/s}\)  and  \(\vec {\text v}_2=8\,\text{m/s}\) respectively.
• According to man \(\text m_1\), he is at rest and \(\text m_2\) is completely responsible for the change in distance between them, as \(\text m_2\) is moving at \(5\,\hat {\text i}\,\text{m/s}\) .

Mathematically,

\(\vec {\text v}_{2/1}=\vec {\text v}_2-\vec {\text v}_1\)

where, 

\(\vec {\text v}_{2/1}\rightarrow\) velocity of m₂ as observed by m₁ 

\(\vec {\text v}_{2}\rightarrow\) velocity of m₂

\(\vec {\text v}_{1}\rightarrow\) velocity of m₁

Velocity of object with respect to observer = Velocity of object being observed – Velocity of observer

\(\text v_{2/1}=8-3=5\,\text{m/s}\) in positive \(\text x\) - direction i.e., \(5\,\hat{\text i}\,\text{m/s}\)

• Consider two cars \(A\) and \(B\), moving along the same straight line.
• At particular instant their velocities, positions and accelerations are shown in the figure.

• To calculate time of collision:

Step1: Write down the position vector of car \(A\) and car \(B\) as a function of time.

Step2 : After equating them, we obtained the value of \(t\) which is the time when cars will collide.

• Another method is the use of relative motion.

1.  Attach frame of reference on one of the cars.

2.  At the time of collision, their relative difference will reduce to zero.

3.  Let car \(A\) is taken as frame of reference.

• Position vector, \(\vec x_{B/A}=(x_B-x_A)\,\hat i\)

where, \(\vec x_{B/A}\) is a position vector of car \(B\) with respect to car \(A\)

• Velocity vector,  \(\vec {\text v}_{B/A}=(\text v_B-\text v_A)\,\hat i\)

where, \(\vec {\text v}_{B/A}=\) velocity of \(B\) with respect to \(A\)

• Acceleration vector,  \(\vec a_{B/A}=(a_B-a_A)\,\hat i\)

where, \(\vec a_{B/A}=\) acceleration of \(B\) with respect to \(A\)

• Using \((s)_t=s_0+ut+\dfrac{1}{2}at^2\)

\((\vec x_{B/A})_t=\vec x_{B/A}+(\text v_B-\text v_A)\,t\;\hat i+\dfrac{1}{2}(a_B-a_A)\,t^2\,\hat i\)

• At the time of collision,

\((\vec x_{B/A})_t=0\)

So,

\(0=\vec x_{B/A}+(\text v_B-\text v_A)\,t\;\hat i+\dfrac{1}{2}(a_B-a_A)\,t^2\,\hat i\)

\(t=\,\,\dfrac{-(\text v_B-\text v_A)\pm\sqrt{(\text v_B-\text v_A)^2-4×\dfrac{1}{2}(a_B-a_A)\,x_{B/A}}}{2×\dfrac{1}{2}(a_B-a_A)}\)

\(t=\,\,\,\dfrac{-(\text v_B-\text v_A)\pm\sqrt{(\text v_B-\text v_A)^2-2\,(a_B-a_A)\,x_{B/A}}}{a_B-a_A}\)

• Consider a situation in which a passenger is moving in straight line with constant velocity \(\text v_0\hat {\text i}\) on the platform.
• A train is also moving with constant velocity  \(\text v_{\text t} \hat {\text i}\), as shown in figure. 
• The velocities shown are in reference frame attached to the ground.
• Now, the reference frame is attached to the passenger. He observes that the velocity of train is not same as before.
• Because, according to passenger, his velocity is zero.
• To make him at rest in frame of reference attached to him, add  \(-\vec{\text v}_0\)  to all objects.

\(\vec {\text v}_0=\text v_0\hat {\text i}=\) velocity of passenger in ground frame

\(\vec {\text v}_{\text t}= {\text v}_{\text t}\hat {\text i}=\) velocity of train in ground frame

\(\vec {\text v}_{{\text t}/0}=\) velocity of train observed by passenger

\(\vec {\text v}_{{\text t}/0}=\vec{\text v}_{\text t}-\vec {\text v}_0\)

\(\vec {\text v}_{{\text t}/0}={\text v}_{\text t}\hat {\text i}- {\text v}_0\hat {\text i}\)

Relative Motion in One-Dimension

• Two men \(\text m_1\) and \(\text m_2\) are moving along the same line, as shown in figure.
• They both have different velocities, \(\vec {\text v}_1=3 \,\text {m/s}\)  and  \(\vec {\text v}_2=8\,\text{m/s}\) respectively.
• According to man \(\text m_1\), he is at rest and \(\text m_2\) is completely responsible for the change in distance between them, as \(\text m_2\) is moving at \(5\,\hat {\text i}\,\text{m/s}\).

Mathematically,

\(\vec {\text v}_{2/1}=\vec {\text v}_2-\vec {\text v}_1\)

where, 

\(\vec {\text v}_{2/1}\rightarrow\) velocity of m₂ as observed by m₁ 

\(\vec {\text v}_{2}\rightarrow\) velocity of m₂

\(\vec {\text v}_{1}\rightarrow\) velocity of m₁

Velocity of object with respect to observer = Velocity of object being observed – Velocity of observer

\(\text v_{2/1}=8-3=5\,\text{m/s}\) in positive \(\text x\) - direction i.e., \(5\,\hat{\text i}\,\text{m/s}\) 

Relative Acceleration

\(\vec {\text v}_{{\text t}/0}=\vec{ \text v}_\text t-\vec {\text v}_0\)

Differentiating with respect to time

\(\vec {\text a}_\text{t/0}=\vec {\text a}_\text t-\vec {\text a}_\text 0\)

where,  \(\vec {\text a}_{\text t/0}=\) acceleration of train in the frame of reference of observer

\(\vec {\text a}_{\text t}=\) acceleration of train in ground frame

\(\vec {\text a}_{0}=\) acceleration of observer in ground frame

• Consider, two particles A and B projected vertically upwards under gravity then it moves with constant acceleration.
• This acceleration is \(g\) i.e., acceleration due to gravity.
• Since both the particles are projected under gravity so, their relative acceleration under gravity is zero. 
• To calculate the time when the particle will meet use,

\(s_f=s_i+\text v_{A/B}\,t+\dfrac{1}{2}a_{A/B}\,t^2\)

where, 

\(\text{v}_{A/B}\) is the relative velocity of A with respect to B

\(a_{A/B}\) is the relative acceleration of A with respect to B

• Since relative acceleration is zero because both motions are under gravity so, \(a_{A/B}=0\).

\(s_f=s_i+\text v_{A/B}\,t\)                      -(1)

• Relative velocity of \(A\) with respect to \(B\) considering \(A\) as reference frame,

  \(\text v_{A/B}=\text v_A-\text v_B\)

• Also when the particles meet \(s_f=0\)
• So bt putting values in equation we get the time when the particle meet.

• Consider a situation in which a passenger is moving in straight line with constant velocity \(\text v_0\hat i\) on the platform.
• A train is also moving with constant velocity \(\text v_t \hat i\), as shown in figure. 
• The velocities shown are in reference frame attached to the ground.
• Now, the reference frame is attached to the passenger. He observes that the velocity of train is not same as before.
• Because, according to passenger, his velocity is zero.
• To make him at rest in frame of reference attached to him, add \(-\vec {\text v}_0\) to all objects.

• \(\vec {\text v}_0=\text v_0\hat i=\) velocity of passenger in ground frame

  \(\vec {\text v}_t=\vec {\text v}_t\hat i=\) velocity of train in ground frame

  \(\vec {\text v}_{t/0}=\) velocity of train observed by passenger

  \(\vec {\text v}_{t/0}=\vec {\text v}_t-\vec {\text v}_0\)

  \(\vec {\text v}_{t/0}=\vec {\text v}_t\hat i-\vec {\text v}_0\hat i\)

   

Relation Motion in One-Dimension 

• Two men \(m_1\) and \(m_2\) are moving along the same line, as shown in figure.
• They both have different velocities, \(\vec {\text v}_1=3m/s\) and \(\vec {\text v}_2=8m/s\) respectively.
• According to man \(m_1\), he is at rest and \(m_2\) is completely responsible for the change in distance between them, as \(m_2\) is moving at \(5\,\hat {\text i}\,m/s\).

• Mathematically,

  \(\vec {\text v}_{2/1}=\vec {\text v}_2-\vec {\text v}_1\)

  where, 

  \(\vec {\text v}_{2/1}\rightarrow\) Velocity of m₂ as observed by m₁ 

  \(\vec {\text v}_{2}\rightarrow\) Velocity of m₂

  \(\vec {\text v}_{1}\rightarrow\) Velocity of m₁

  Velocity of object with respect to observer = velocity of object being observed – velocity of observer

  \(\text v_{2/1}=8-3=5\,m/s\) in positive \(\text x\)- direction i.e., \(5\,\hat i\,m/s\)

   

Relative Acceleration

\(\vec {\text v}_{t/0}=\vec {\text v}_t-\vec {\text v}_0\)

Differentiating with respect to time

\(\vec a_{t/0}=\vec a_t-\vec a_0\)

where, \(\vec a_{t/0}=\) acceleration of train in the frame of reference of observer

\(\vec a\,_t=\) acceleration of train in ground frame

\(\vec a_0=\) acceleration of observer in ground frame