Informative line

Problems Involving Two Dimensional Motion

Learn relative motion examples in two dimensional. Practice problems of kinematics in two dimensional motion and vectors.

Relative Velocity

• Consider a situation in which a passenger is moving in straight line with constant velocity $$\text v_o\hat {\text i}$$ 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 x - direction i.e., $$5\,\hat{\text i}\,\text{m/s}$$

Similarly,

• Consider a particle is located at point B and another particle is located at point A, as shown in figure.
• Let  $$\vec{r}_{B}$$ = Position vector of B with respect to origin or ground

$$\vec{r}_A$$ = Position vector of A with respect to origin or ground

$$\vec{r}_{B/A}$$ = Position vector of B with respect to observer A

$$\vec{r}_A+\vec{r}_{B/A}=\vec{r}_B$$

$$\Rightarrow\vec{r}_{B/A}=\vec{r}_B-\vec{r}_A$$

After differentiating both sides

$$\Rightarrow \vec{v}_{B/A}=\vec{v}_B-\vec{v}_A$$

Consider two cars A and B, moving with a speed of 40 m/s towards east and 30 m/s towards north respectively. Determine the direction of velocity of A with respect to the direction of velocity of B.

A $$tan^{-1}\left(\dfrac{3}{4}\right)$$ South of East

B $$tan^{-1}\left(\dfrac{3}{4}\right)$$ North of East

C $$tan^{-1}\left(\dfrac{4}{3}\right)$$ North of West

D $$tan^{-1}\left(\dfrac{4}{3}\right)$$ North of East

×

Velocity,

$$\vec{v}_A=40 \,\hat i\;m/s$$

$$\vec{v}_B=30\, \hat j\;m/s$$

Take B, as reference frame,

$$\vec{v}_{A/B}$$ = Relative velocity of A with respect to B

$$\vec {v}_{A/B}=\vec{v}_A-\vec{v}_B=\vec{v}_A+(-\vec {v}_B)$$

$$=40 \hat i -30 \hat j$$

Here

$$\vec{v}_{A/B}=\vec{v}_A-\vec{v}_B$$

(Velocity of A as seen by B)

Here, $$tan\,\theta=\dfrac{30}{40}$$

$$\theta=tan^{-1}\left(\dfrac{3}{4}\right)$$

So A will be moving $$tan^{-1}\left(\dfrac{3}{4}\right)$$ south of east with respect to B

Consider two cars A and B, moving with a speed of 40 m/s towards east and 30 m/s towards north respectively. Determine the direction of velocity of A with respect to the direction of velocity of B.

A

$$tan^{-1}\left(\dfrac{3}{4}\right)$$ South of East

.

B

$$tan^{-1}\left(\dfrac{3}{4}\right)$$ North of East

C

$$tan^{-1}\left(\dfrac{4}{3}\right)$$ North of West

D

$$tan^{-1}\left(\dfrac{4}{3}\right)$$ North of East

Option A is Correct

The Man and the Rain Problem

• A man is moving with a velocity $$\vec{v}_m$$ along x-axis.
• Rain is falling vertically downwards with a velocity $$\vec{v}_r$$.

• Due to relative motion, man observes that rain is falling at some angle $$\theta$$ with the vertical direction.

$$\vec{v}_{r/m}$$ = Velocity of rain with respect to man

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$tan\,\theta=\dfrac{|\vec{v}_m|}{|\vec{v}_r|}$$

$$\theta=tan^{-1}\dfrac{|\vec{v}_m|}{|\vec{v}_r|}$$

• Rain is falling at an angle $$\theta=tan^{-1}\dfrac{|\vec{v}_m|}{|\vec{v}_r|}$$with the vertical direction.

A man is moving with a velocity $$(\vec v_m)$$ of $$4\,m/s$$ towards x-axis. If rain is falling vertically downwards with a velocity $$(\vec v_r)$$ of $$3\,m/s$$, then at which angle should he keep his umbrella to avoid the rain?

A $$\theta=tan^{-1}\left(\dfrac{2}{5}\right)$$

B $$\theta=tan^{-1}\left(\dfrac{3}{2}\right)$$

C $$\theta=tan^{-1}\left(\dfrac{5}{4}\right)$$

D $$\theta=tan^{-1}\left(\dfrac{4}{3}\right)$$

×

Take man as reference frame,

$$\vec{v}_{r/m}$$ = Relative velocity of rain with respect to man

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$=(-3 \hat j -4 \hat i) \;m/s$$

Here

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

(Velocity of rain as seen by man)

$$tan \,\theta=\dfrac{|\vec{v}_m|}{|\vec{v}_r|}=\dfrac{4}{3}$$

$$\theta=tan^{-1}\left(\dfrac{4}{3}\right)$$

So, he should keep his umbrella at an angle $$\theta=tan^{-1}\left(\dfrac{4}{3}\right)$$ with the vertical to avoid the rain.

A man is moving with a velocity $$(\vec v_m)$$ of $$4\,m/s$$ towards x-axis. If rain is falling vertically downwards with a velocity $$(\vec v_r)$$ of $$3\,m/s$$, then at which angle should he keep his umbrella to avoid the rain?

A

$$\theta=tan^{-1}\left(\dfrac{2}{5}\right)$$

.

B

$$\theta=tan^{-1}\left(\dfrac{3}{2}\right)$$

C

$$\theta=tan^{-1}\left(\dfrac{5}{4}\right)$$

D

$$\theta=tan^{-1}\left(\dfrac{4}{3}\right)$$

Option D is Correct

Calculation of Angle of Rain Falling

• Consider a man is running along x-axis.
• Rain is falling at an angle $$\phi$$ with the vertical direction.

• Due to relative motion, man observes that rain is falling at some different angle $$\theta$$ with the vertical direction.

$$\vec{v}_m$$ = velocity of man

$$\vec{v}_r$$ = velocity of rain

$$\vec{v}_{r/m}$$ = velocity of rain with respect to man

Calculation of $$\theta$$

• Assuming horizontal direction as x-axis and vertical direction as y-axis.
• Resolve $$\vec{v}_r$$ and $$\vec{v}_m$$ in component form.
• Calculate $$\vec{v}_{r/m}$$,   $$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$
• Angle $$tan\,\theta=\dfrac{|(v_{r/m})_x|}{|(v_{r/m})_y|}$$

A man is running horizontally with a velocity of 8 m/s towards right. Rain is falling with a velocity of 5 m/s making an angle 37º with the vertical direction. Calculate the angle with direction of velocity of rain with respect to man. $$\Big[Take\, sin\,37º =\dfrac{3}{5} \,\,\,cos \,37^o=\dfrac{4}{5}\Big]$$

A $$tan^{-1}(3)$$

B $$tan^{-1}\left(\dfrac{5}{4}\right)$$

C $$tan^{-1}\left(\dfrac{11}{4}\right)$$

D $$tan^{-1}(2)$$

×

$$\vec{v}_m=8 \hat i\; m/s$$

$$\vec{v}_r$$ making an angle 37º with the vertical direction .

$$\vec v_r=(-5\,sin\,37^o) \hat i -(5 \,cos \,37^o) \hat j$$

$$\Big(-5×\dfrac {3}{5}\Big)\hat i-\Big(5×\dfrac{4}{5}\Big)\hat j$$

So, $$\vec{v}_r=(-3 \hat i -4 \hat j) \;m/s$$

$$\vec{v}_{r/m}$$ = Relative velocity of rain with respect to man

as   $$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$=(-3 \hat i -4 \hat j) -8 \hat i$$

$$=(-11 \hat i -4 \hat j) \;m/s$$

Here,  $$tan\,\theta=\dfrac{|(v_{r/m})_x|}{|(v_{r/m})_y|}$$

$$tan\,\theta=\dfrac{|-11|}{|-4|}$$

$$\theta=tan^{-1}\left(\dfrac{11}{4}\right)$$

A man is running horizontally with a velocity of 8 m/s towards right. Rain is falling with a velocity of 5 m/s making an angle 37º with the vertical direction. Calculate the angle with direction of velocity of rain with respect to man. $$\Big[Take\, sin\,37º =\dfrac{3}{5} \,\,\,cos \,37^o=\dfrac{4}{5}\Big]$$

A

$$tan^{-1}(3)$$

.

B

$$tan^{-1}\left(\dfrac{5}{4}\right)$$

C

$$tan^{-1}\left(\dfrac{11}{4}\right)$$

D

$$tan^{-1}(2)$$

Option C is Correct

Condition for Collision

• Consider two particles, projected from point A and point B, as shown in figure.
• These particles will collide, if the relative velocity of one with respect to another is towards itself.
• Taking A as a reference frame.
• Relative position vector of B with respect to A

$$\vec{r}_{B/A}=\vec{r}_B-\vec{r}_A$$

$$\hat{r}_{B/A}=\dfrac{\vec{r}_B-\vec{r}_A}{|\vec r_ B-\vec{r}_A|}$$

• Relative velocity of B with respect to A

$$\vec{v}_{B/A}=\vec{v}_B-\vec{v}_A$$

$$\hat{v}_{B/A}=\dfrac{\vec{v}_B-\vec{v}_A}{|\vec{v}_B-\vec{v}_A|}$$

• If they collide, both unit vectors must be towards each other.

$$\hat{r}_{B/A}= -\hat{v}_{B/A}$$

$$\dfrac{\vec{r}_B-\vec{r}_A}{|\vec{r}_B-\vec{r}_A|} =-\left(\dfrac{\vec{v}_B-\vec{v}_A}{|\vec{v}_B-\vec{v}_A|}\right)$$

Particle A is projected with a velocity of $$v_A=5\,m/s$$, making an angle $$\theta$$ with the horizontal direction. Second particle B is projected vertically upward with a velocity $$v_B=3\,m/s$$, as shown in figure. At which angle $$\theta$$, A should be projected so that A and B will collide?

A $$\theta=sin^{-1}\left(\dfrac{4}{5}\right)$$

B $$\theta=sin^{-1}\left(\dfrac{3}{5}\right)$$

C $$\theta=sin^{-1}\left(\dfrac{3}{7}\right)$$

D $$\theta=cos^{-1}\left(\dfrac{3}{5}\right)$$

×

Velocity of particle A, $$v_A=5\,m/s$$

Velocity of particle B, $$v_B=3\,m/s$$

B as a reference frame

For collision, direction of velocity A must be towards B.

$$5\sin\theta-3=0$$

$$\sin\theta=\dfrac{3}{5}$$

$$\theta=sin^{-1}\left(\dfrac{3}{5}\right)$$

Particle A is projected with a velocity of $$v_A=5\,m/s$$, making an angle $$\theta$$ with the horizontal direction. Second particle B is projected vertically upward with a velocity $$v_B=3\,m/s$$, as shown in figure. At which angle $$\theta$$, A should be projected so that A and B will collide?

A

$$\theta=sin^{-1}\left(\dfrac{4}{5}\right)$$

.

B

$$\theta=sin^{-1}\left(\dfrac{3}{5}\right)$$

C

$$\theta=sin^{-1}\left(\dfrac{3}{7}\right)$$

D

$$\theta=cos^{-1}\left(\dfrac{3}{5}\right)$$

Option B is Correct

Velocity of Man such that Rain Appears Vertically downwards to him

• Consider a man is running on a horizontal road with some velocity.
• Rain is falling with velocity $$v_r$$, making an angle $$\theta$$ with the vertical direction.
• The rain appears falling vertically to the man.
• Let the velocity of man be $$\vec{v}_m= v_{m} \hat i$$ along x-axis.
• Let $$\vec{v}_r= -a \hat i-b\hat j$$  where $$\dfrac{b}{a}=tan\theta$$
• Let $$\vec{v}_{r/m}$$ = Velocity of rain with respect to man
• For $$\vec{v}_{r/m}$$, rain appears vertically to the man, $$(v_{r/m})_x=0$$

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$=-a \hat i-b\hat j-v_{m} \hat i$$

$$=(-a -v_m)\hat i-b\hat j$$

$$\because(v_{r/m})_x=0$$

$$\Rightarrow -a -v_m = 0$$

$$\Rightarrow v_m=-a$$

$$v_m=-a\hat i$$

A man is running on a horizontal road. Rain is falling with a velocity of 5 m/s making an angle $$\theta=37^o$$ with the vertical, but man observes that rain is falling vertically downwards. Determine the velocity of man. $$\Big[Take \,sin 37º = \dfrac{3}{5}\Big]$$

A $$-5\;m/s$$ toward left

B $$-4\;m/s$$ toward left

C $$3\;m/s$$ toward left

D $$-2\;m/s$$ toward left

×

Let $$\vec{v}_m= v \hat i$$ m/s

$$\vec{v}_r$$ is making an angle $$\theta=37^o$$ with the vertical direction

So,  $$\vec{v}_r=(-3 \hat i -4 \hat j) \;m/s$$

$$\vec{v}_{r/m}$$ = Velocity of rain with respect to man

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$=(-3 \hat i -4 \hat j) -v\hat i$$

$$\vec{v}_{r/m}=[(-3-v)\hat i-4 \hat j]\,m/s$$

$$=[(v_{r/m})_x \hat i+(v_{r/m})_y \hat j]$$

Here,

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$=\vec{v}_r-\vec{v}$$

Man observes that rain is falling vertically downward

So, $$(v_{r/m})_x=0$$

$$-3-v=0$$

$$v=-3\,m/s$$

So, man is running with a velocity 3 m/s towards left.

A man is running on a horizontal road. Rain is falling with a velocity of 5 m/s making an angle $$\theta=37^o$$ with the vertical, but man observes that rain is falling vertically downwards. Determine the velocity of man. $$\Big[Take \,sin 37º = \dfrac{3}{5}\Big]$$

A

$$-5\;m/s$$ toward left

.

B

$$-4\;m/s$$ toward left

C

$$3\;m/s$$ toward left

D

$$-2\;m/s$$ toward left

Option C is Correct

Relative Trajectory under Constant Acceleration

• Consider a particle is projected from point A, and other particle is projected from point B, as shown in figure.

• The relative acceleration of particle A with respect to particle B is zero.
• Let $$a_{A/B}$$ = Relative acceleration of A with respect to B

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

$$a_{A/B}=g-g$$

$$a_{A/B}=0$$

• Motion of particle under zero acceleration is a straight line.

Two particles A and B are projected with initial velocities $$3\,m/s$$ and $$5\,m/s$$, as shown in figure. What will be the trajectory of particle A as observed by B?

A Parabola

B Straight Line

C Circular Path

D Triangular Path

×

As both particles are projected under gravity, the relative acceleration will be zero.

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

$$=g-g=0$$

where $$a_{A/B}$$ = acceleration of A with respect to B

Under gravity, relative trajectory of particles is straight line.

Two particles A and B are projected with initial velocities $$3\,m/s$$ and $$5\,m/s$$, as shown in figure. What will be the trajectory of particle A as observed by B?

A

Parabola

.

B

Straight Line

C

Circular Path

D

Triangular Path

Option B is Correct

Shortest Distance Between Two Moving Bodies

• Consider two bodies A and B are d distance apart from each other.

Let

$$\vec{v}_A$$ = Velocity of A

$$\vec{v}_B$$ = Velocity of B

$$\vec{v}_{A/B}$$ = Velocity of A with respect to B

If bodies A and B do not collide, then the minimum separation can be calculated as follows:

• According to the reference frame of B, the minimum separation will be perpendicular to $$B_0$$ along the direction of $$\vec{v}_{A/B}$$.

$$B_0=d\;sin\phi$$

$$tan\;\phi=\dfrac{|(v_{A/B})_y|}{|(v_{A/B})_x|}$$

Two cars A and B are moving with velocities $$v_A=20\,m/s$$ and $$v_B=15\,m/s$$ respectively, as shown in figure. Calculate the minimum distance between the cars if they do not collide.

A $$16\,m$$

B $$20\,m$$

C $$10\,m$$

D $$15\,m$$

×

According to ground frame

Velocity of car A = 20 m/s

Velocity of car B = 15 m/s

Car B is taken as reference frame and path of car A is seen from car B.

Minimum distance between cars will be the perpendicular distance from car B on the path of car A as observed by car B.

If these two cars do not collide,

BD = Minimum separation

Here

$$tan\,\theta=\dfrac{|\vec{v}_B|}{|\vec{v}_A|}$$

$$tan\,\theta=\dfrac{15}{20}$$

$$tan\theta=\dfrac{3}{4}$$

$$\therefore sin\,\theta=\dfrac{3}{5}\;\;,\;\;cos\,\theta=\dfrac{4}{5}$$

Distance, $$OC=AO \;tan\,\theta$$

$$=500×\dfrac{3}{4} =375\,m$$

Distance, $$CB=OB-OC$$

$$=400-375=25 \,m$$

Distance, $$BD=CB\;cos\theta$$

$$=25×\dfrac{4}{5} =20\,m$$

Two cars A and B are moving with velocities $$v_A=20\,m/s$$ and $$v_B=15\,m/s$$ respectively, as shown in figure. Calculate the minimum distance between the cars if they do not collide.

A

$$16\,m$$

.

B

$$20\,m$$

C

$$10\,m$$

D

$$15\,m$$

Option B is Correct

Speed of Rain with Respect to Man Running Horizontally

• A man is running with a velocity $$v_m$$ along x-axis.
• The rain is falling with velocity $$\vec{v}_r$$ making an angle $$\theta$$ with the vertical direction as shown in figure.

$$\vec{v}_r=-(v_r\;sin\theta)\hat i -(v_r\;cos\theta)\hat j$$

• Let $$\vec{v}_{r/m}$$ = velocity of rain with respect to man

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$=-(v_r\;sin\theta)\hat i -(v_r\;cos\theta)\hat j-v_m \hat i$$

$$=-(v_m+v_r\;sin\theta)\hat i -(v_r\;cos\theta)\hat j$$

$$|\vec{v}_{r/m}|=\sqrt{(v_m+v_rsin\theta)^2+(v_r\,cos\theta)^2}$$

$$=\sqrt{v_m^2+v_r^2\;sin^2\theta+v_r^2\;cos^2\theta+2\,v_mv_r\sin\theta}$$

$$|\vec{v}_{r/m}|=\sqrt{v_m^2+v_r^2\;+2\,v_mv_r\sin\theta}$$

A man is running with a velocity $$(\vec v_m)$$ of $$5\,m/s$$ towards right and the rain is falling with a velocity $$(v_r)$$ of $$5\,m/s$$ making an angle $$\theta=37^o$$ with the vertical direction. Calculate the speed of rain with respect to man.

A $$3\sqrt{7}\;ms^{-1}$$

B $$4\sqrt{5}\;ms^{-1}$$

C $$5\sqrt{6}\;ms^{-1}$$

D $$3\sqrt{5}\;ms^{-1}$$

×

$$\vec{v}_m=5\hat i\;m/s$$

$$\vec{v}_r$$ is making angle 37º with the vertical direction

$$\vec{v}_r = -5\,sin\;37^o\hat i - 5\,cos\;37^o\hat j$$

$$=(-3\hat i - 4\hat j)\,m/s$$

Taking man as reference frame,

$$\vec{v}_{r/m}$$ = velocity of rain with respect to man

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

$$=(-3\hat i - 4\hat j)-5\hat i$$

$$=(-8\hat i - 4\hat j)\;m/s$$

Here

$$\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m$$

(Velocity of rain as seen by man)

Speed of rain with respect to man = $$|\vec{v}_{r/m}|$$

$$|\vec{v}_{r/m}|=\sqrt{(8)^2+(4)^2}$$

$$=\sqrt{64+16}$$

$$=\sqrt{80}$$

$$=4\sqrt{5}\;ms^{-1}$$

A man is running with a velocity $$(\vec v_m)$$ of $$5\,m/s$$ towards right and the rain is falling with a velocity $$(v_r)$$ of $$5\,m/s$$ making an angle $$\theta=37^o$$ with the vertical direction. Calculate the speed of rain with respect to man.

A

$$3\sqrt{7}\;ms^{-1}$$

.

B

$$4\sqrt{5}\;ms^{-1}$$

C

$$5\sqrt{6}\;ms^{-1}$$

D

$$3\sqrt{5}\;ms^{-1}$$

Option B is Correct