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Problems Involving One Dimensional Motion

Learn relative motion examples in one dimensional and relative velocity & acceleration. Practice problems to calculation of relative velocity and relative acceleration.

Idea of Relative Motion

• 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.

Alia is $$5\,ft$$ tall and Sara is $$4\,ft$$ tall. What is the relative height of Alia with respect to Sara?

A $$1\,ft$$

B $$4\,ft$$

C $$5\,ft$$

D $$6\,ft$$

×

Relative height of Alia with respect to Sara is given as,

Relative Height = Height of Alia – Height of Sara

$$=5ft-4ft$$

$$=1ft$$

Hence, option $$A$$ is correct.

Alia is $$5\,ft$$ tall and Sara is $$4\,ft$$ tall. What is the relative height of Alia with respect to Sara?

A

$$1\,ft$$

.

B

$$4\,ft$$

C

$$5\,ft$$

D

$$6\,ft$$

Option A is Correct

Relative Velocity

• 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 a situation in which A is moving with a speed of 5 m/s towards right. B is moving with a speed of 8 m/s towards left. What is the velocity of B relative to A?

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 $$14\, \hat{\text i}\,\text{m/s}$$

×

$$\vec{\text v}_\text A=5$$ $$\hat{\text i}\,\text{m/s}$$

$$\vec {\text v}_\text B=-8$$ $$\hat{\text i}\,\text{m/s}$$ A is taken as frame of reference,

$$\vec {\text v}_\text {B/A}=$$ velocity of B as seen by A

$$\vec {\text v}_\text{B/A}=\vec {\text v}_\text B-\vec{\text v}_\text A$$ $$\vec{\text v}_\text {B/A}=-8\hat {\text i}-5\hat {\text i}$$

$$=\, –13\,\hat{\text i}\,\text{m/s}$$ For $$A,\;B$$ will be coming towards him with a speed of  $$13 \,\text{m/s}$$. Consider a situation in which A is moving with a speed of 5 m/s towards right. B is moving with a speed of 8 m/s towards left. What is the velocity of B relative to A? 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

$$14\, \hat{\text i}\,\text{m/s}$$

Option C is Correct

Time of Collision

• 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,

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

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

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

Two cars start moving in same direction with a distance, $$s_i=4\,m$$ between them. The initial velocities of cars, $$\text v_A=8\,m/s$$ and $$\text v_B=4\,m/s$$ and accelerations, $$a_A=2\,m/s^2$$ and $$a_B=4\,m/s^2.$$ Find the time when they will meet.

A $$5\,sec$$

B $$4\,sec$$

C $$10\,sec$$

D $$2\,sec$$

×

Let car $$A$$ is taken as frame of reference.

Relative velocity of car $$B$$ with respect to car $$A$$

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

$$=4-8$$

$$=-4\,m/s$$

Relative acceleration of car $$B$$ with respect to car $$A$$

$$a_{B/A}=a_B-a_A$$

$$=4-2$$

$$=2\,m/s^2$$

Initial separation between cars, $$(s_i)=4\,m$$

Final relative separation between cars, $$(s_f)=0$$

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

$$0=4-4t+\dfrac{1}{2}×2\,t^2$$

$$0=4-4t+t^2$$

$$t=2\,sec$$

Two cars start moving in same direction with a distance, $$s_i=4\,m$$ between them. The initial velocities of cars, $$\text v_A=8\,m/s$$ and $$\text v_B=4\,m/s$$ and accelerations, $$a_A=2\,m/s^2$$ and $$a_B=4\,m/s^2.$$ Find the time when they will meet. A

$$5\,sec$$

.

B

$$4\,sec$$

C

$$10\,sec$$

D

$$2\,sec$$

Option D is Correct

Two cars start moving towards each other when initial separation between them is $$6\,m$$. If initial velocity of both the cars is $$2\,m/s$$ and acceleration is $$2\,m/sec^2$$, find the time when they will collide.

A $$2\,sec$$

B $$3\,sec$$

C $$1\,sec$$

D $$4\,sec$$

×

Let car $$A$$ is taken as frame of reference.

Relative velocity of car $$B$$ with respect to $$A$$

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

$$=-2-2$$

$$=-4\,m/s$$

(Assuming right as positive direction)

Relative acceleration of car $$B$$ with respect to $$A$$

$$a_{B/A}=-a_B-(+a_A)$$

$$=-2-2$$

$$=-4\,m/s^2$$

Initial separation, $$(s_i)=6\,m$$

Final separation, $$(s_f)=0$$

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

$$0=6-4t-\dfrac{1}{2}×4\,t^2$$

$$=6-4t-2t^2$$

$$0=3-2t-t^2$$

$$t=1\,sec$$

Two cars start moving towards each other when initial separation between them is $$6\,m$$. If initial velocity of both the cars is $$2\,m/s$$ and acceleration is $$2\,m/sec^2$$, find the time when they will collide. A

$$2\,sec$$

.

B

$$3\,sec$$

C

$$1\,sec$$

D

$$4\,sec$$

Option C is Correct

Relative Acceleration

• 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

Alia(A) and Riya(R) are moving with acceleration 8 m/s2 and 3 m/s2 respectively, along $$x$$-axis. Calculate the relative acceleration of R with respect to A.

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 $$– 6$$ $$\hat {\text i}\,\text{m/sec}^2$$

×

Acceleration, $$\vec {\text a}_\text A=8$$ $$\hat {\text i}\,\text{m/sec}^2$$

$$\vec {\text a}_\text R=3$$ $$\hat {\text i}\,\text{m/sec}^2$$ A is taken as frame of reference,

$$\vec {\text a}_\text{R/A}=$$ Acceleration of R with respect to A

$$\vec {\text a}_\text {R/A}=\vec{\text a}_\text R-\vec{\text a}_\text A$$

$$=\left(3\,\hat i-8\,\hat i\right)$$

$$\vec {\text a}_\text{R/A}=-5$$ $$\hat {\text i}\,\text{m/sec}^2$$ Alia(A) and Riya(R) are moving with acceleration 8 m/s2 and 3 m/s2 respectively, along $$x$$-axis. Calculate the relative acceleration of R with respect to A. 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

$$– 6$$ $$\hat {\text i}\,\text{m/sec}^2$$

Option A is Correct

Particle Projected in Same Direction Under Gravity

• 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 particle $$A$$, projected from height $$h=10\,m$$ with initial velocity $$\text v_A=2\,m/s$$. Another particle $$B$$ is projected vertically upwards in same direction with initial velocity $$\text v_B=12\,m/s$$ from ground. Find the time when they will meet.

A $$1\,sec$$

B $$8\,sec$$

C $$5\,sec$$

D $$4\,sec$$

×

Let upward direction be positive direction.

Let particle $$A$$ be taken as frame of reference.

Relative velocity of $$A$$ with respect to $$B$$

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

$$=2-12$$

$$=-10\,m/s$$

Relative acceleration is zero because both motions are under gravity.

Initial separation, $$(s_i)=10\,m$$

Final separation, $$(s_f)=0$$

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

$$0=10-10t+0$$

$$t=1\,sec$$

Consider a particle $$A$$, projected from height $$h=10\,m$$ with initial velocity $$\text v_A=2\,m/s$$. Another particle $$B$$ is projected vertically upwards in same direction with initial velocity $$\text v_B=12\,m/s$$ from ground. Find the time when they will meet. A

$$1\,sec$$

.

B

$$8\,sec$$

C

$$5\,sec$$

D

$$4\,sec$$

Option A is Correct

Consider a particle $$A$$, projected from height $$h=10\,m$$ with initial velocity $$\text u_A=2\,m/s$$, another particle $$B$$ is projected vertically upwards in same direction with initial velocity $$\text u_B=12\,m/s$$ from ground. If they meet after $$1\,sec$$, then find the height from ground at which they collide.

A $$17\,m$$

B $$8\,m$$

C $$7\,m$$

D $$12\,m$$

×

Let $$H$$ be the height, where they collide, which is attained after $$1\,sec$$.

For particle $$B$$

$$H=u_{B}\,t-\dfrac{1}{2}\,g\,t^2$$

$$H=12×1-\dfrac{1}{2}×10×(1)^2$$

$$H=7\,m$$

Consider a particle $$A$$, projected from height $$h=10\,m$$ with initial velocity $$\text u_A=2\,m/s$$, another particle $$B$$ is projected vertically upwards in same direction with initial velocity $$\text u_B=12\,m/s$$ from ground. If they meet after $$1\,sec$$, then find the height from ground at which they collide.

A

$$17\,m$$

.

B

$$8\,m$$

C

$$7\,m$$

D

$$12\,m$$

Option C is Correct

Calculation of Relative Velocity and Relative Acceleration

• 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

Alia and Riya are moving with velocity $$3\,m/sec$$ and $$2\,m/sec$$ respectively, along $$x$$-axis. If their accelerations are $$4\,m/sec^2$$ and $$2\,m/sec^2$$ respectively, along $$x$$-axis then find the relative speed and acceleration of $$R$$ with respect to $$A$$.

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 $$4\,\hat i\,m/s,\,2\,\hat i\,m/s^2$$

×

Let $$A$$ is taken as frame of reference.

Relative velocity of $$R$$ with respect to $$A$$

$$\vec {\text v}_{R/A}=\vec {\text v}_R-\vec {\text v}_A$$

$$\vec {\text v}_{R/A}=+2\,\hat i-3\,\hat i$$

$$=-\hat i\,m/sec$$ Let $$\vec a_{R/A}=$$ Relative acceleration of $$R$$ with respect to $$A$$

$$\vec a_{R/A}=\vec a_R-\vec a_A$$

$$=2\,\hat i-4\,\hat i$$

$$=-2\,\hat i\,m/sec^2$$ Alia and Riya are moving with velocity $$3\,m/sec$$ and $$2\,m/sec$$ respectively, along $$x$$-axis. If their accelerations are $$4\,m/sec^2$$ and $$2\,m/sec^2$$ respectively, along $$x$$-axis then find the relative speed and acceleration of $$R$$ with respect to $$A$$. 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

$$4\,\hat i\,m/s,\,2\,\hat i\,m/s^2$$

Option B is Correct