Learn relative motion examples in one dimensional and relative velocity & acceleration. Practice problems to calculation of relative velocity and relative acceleration.
A \(1\,ft\)
B \(4\,ft\)
C \(5\,ft\)
D \(6\,ft\)
\(\vec {\text v}_0=\text v_0\hat {\text i}=\) 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
\(\vec {\text v}_{{\text t}/0}=\vec{\text v}_{\text t}-\vec {\text v}_0\)
\(\vec {\text v}_{{\text t}/0}={\text v}_{\text t}\hat {\text i}- {\text v}_0\hat {\text i}\)
Mathematically,
\(\vec {\text v}_{2/1}=\vec {\text v}_2-\vec {\text v}_1\)
where,
\(\vec {\text v}_{2/1}\rightarrow\) velocity of m_{2} as observed by m_{1}
\(\vec {\text v}_{2}\rightarrow\) velocity of m_{2}
\(\vec {\text v}_{1}\rightarrow\) velocity of m_{1}
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., \(5\,\hat{\text i}\,\text{m/s}\)
A \(13 \,\hat{\text i}\,\text{m/s}\)
B \(– 14\,\hat{\text i}\,\text{m/s}\)
C \(– 13\,\hat{\text i}\,\text{m/s}\)
D \(14\, \hat{\text i}\,\text{m/s}\)
\(\vec {\text v}_0=\text v_0\hat {\text i}=\) 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
\(\vec {\text v}_{{\text t}/0}=\vec{\text v}_{\text t}-\vec {\text v}_0\)
\(\vec {\text v}_{{\text t}/0}={\text v}_{\text t}\hat {\text i}- {\text v}_0\hat {\text i}\)
Mathematically,
\(\vec {\text v}_{2/1}=\vec {\text v}_2-\vec {\text v}_1\)
where,
\(\vec {\text v}_{2/1}\rightarrow\) velocity of m_{2} as observed by m_{1}
\(\vec {\text v}_{2}\rightarrow\) velocity of m_{2}
\(\vec {\text v}_{1}\rightarrow\) velocity of m_{1}
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., \(5\,\hat{\text i}\,\text{m/s}\)
\(\vec {\text v}_{{\text t}/0}=\vec{ \text v}_\text t-\vec {\text v}_0\)
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
A \(-5\) \(\hat {\text i}\,\text{m/sec}^2\)
B \(6\,\hat {\text i}\,\text{m/sec}^2\)
C \(5\,\hat {\text i}\,\text{m/sec}^2\)
D \(– 6\) \(\hat {\text i}\,\text{m/sec}^2\)
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.
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.
where, \(\vec x_{B/A}\) is a position vector of car \(B\) with respect to car \(A\)
where, \(\vec {\text v}_{B/A}=\) velocity of \(B\) with respect to \(A\)
where, \(\vec a_{B/A}=\) acceleration of \(B\) with respect to \(A\)
\((\vec x_{B/A})_t=\vec x_{B/A}+(\text v_B-\text v_A)\,t\;\hat i+\dfrac{1}{2}(a_B-a_A)\,t^2\,\hat i\)
\((\vec x_{B/A})_t=0\)
So,
\(0=\vec x_{B/A}+(\text v_B-\text v_A)\,t\;\hat i+\dfrac{1}{2}(a_B-a_A)\,t^2\,\hat i\)
\(t=\,\,\dfrac{-(\text v_B-\text v_A)\pm\sqrt{(\text v_B-\text v_A)^2-4×\dfrac{1}{2}(a_B-a_A)\,x_{B/A}}}{2×\dfrac{1}{2}(a_B-a_A)}\)
\(t=\,\,\,\dfrac{-(\text v_B-\text v_A)\pm\sqrt{(\text v_B-\text v_A)^2-2\,(a_B-a_A)\,x_{B/A}}}{a_B-a_A}\)
A \(5\,sec\)
B \(4\,sec\)
C \(10\,sec\)
D \(2\,sec\)
\(s_f=s_i+\text v_{A/B}\,t+\dfrac{1}{2}a_{A/B}\,t^2\)
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
\(s_f=s_i+\text v_{A/B}\,t\) -(1)
Relative velocity of \(A\) with respect to \(B\) considering \(A\) as reference frame,
\(\text v_{A/B}=\text v_A-\text v_B\)
A \(1\,sec\)
B \(8\,sec\)
C \(5\,sec\)
D \(4\,sec\)
A \(17\,m\)
B \(8\,m\)
C \(7\,m\)
D \(12\,m\)
\(\vec {\text v}_0=\text v_0\hat i=\) 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
\(\vec {\text v}_{t/0}=\vec {\text v}_t-\vec {\text v}_0\)
\(\vec {\text v}_{t/0}=\vec {\text v}_t\hat i-\vec {\text v}_0\hat i\)
Mathematically,
\(\vec {\text v}_{2/1}=\vec {\text v}_2-\vec {\text v}_1\)
where,
\(\vec {\text v}_{2/1}\rightarrow\) Velocity of m_{2} as observed by m_{1}
\(\vec {\text v}_{2}\rightarrow\) Velocity of m_{2}
\(\vec {\text v}_{1}\rightarrow\) Velocity of m_{1}
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., \(5\,\hat i\,m/s\)
\(\vec {\text v}_{t/0}=\vec {\text v}_t-\vec {\text v}_0\)
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
A \(-\hat i\,m/s,\,-4\,\hat i\,m/s^2\)
B \(-\hat i\,m/s,\,-2\,\hat i\,m/s^2\)
C \(-\hat i\,m/s,\,+4\,\hat i\,m/s^2\)
D \(4\,\hat i\,m/s,\,2\,\hat i\,m/s^2\)