Kinematics2

obstacle.

example

Projectile clearing multiple obstacles

Example:
If a projectile crosses two walls of equal height

h

symmetrically as shown in the figure. Find time of flight, height of each wall and maximum height of projectile.

(g = 10 m / s^{2})

Solution:
Since the projectile motion is symmetric,

∴

Total time taken is,

T = 2 + 6 = 8 s

Also,

T = \frac{2 v \sin θ}{g}

and

H = \frac{v^{2} \sin^{2} θ}{2 g}

∴ H = \frac{g}{8} \frac{4 \times v^{2} \sin^{2} θ}{g^{2}}

∴ H = \frac{g \times T^{2}}{8} = \frac{10 \times 8^{2}}{8} = 80

mAlso, for

t = 2 s

assuming height of wall is

h

h = u \sin θ t - \frac{1}{2} g t^{2}

∴ h = 2 u \sin θ - 20

.....(i)
and,

2 u \cos θ = s

.....(ii)
for

t = 6 s

h = + 6 u \sin θ - 180

.....(iii)
and,

6 u \cos θ = s + 120

....(iv)
Equating (i) and (iii),

2 u \sin θ - 20 - 6 u \sin θ + 180 = 0

u \sin θ = \frac{160}{4} = 40

Put in equation (iii)

h = 60 m

example

Projectile on an inclined Plane

An inclined plane is making an angle

β

with horizontal. A projectile is projected from the bottom of the plane with a speed

u

at an angle

α

with horizontal then its range

R

when it strikes the inclined plane is:
Time of flight,

O = u \sin (α - β) t - \frac{1}{2} g \cos β t^{2}

∴ t = \frac{2 u s i n (α - β)}{g c o s β}

Range

O B = (u c o s α) t = \frac{2 u^{2} s i n (α - β) c o s α}{g c o s β}

∴ O A = O B / c o s β = \frac{2 u^{2} s i n (α - β) c o s α}{g c o s^{2} β}

definition

Relative Displacement

Relative displacement, which is displacement of a point on a structure with respect to its original location or an adjacent point on the structure that has also undergone movement, can be an effective indicator of post event structural damage.

Relative displacement is

\vec{r} = {\vec{r}}_{1} - {\vec{r}}_{2}

definition

Relative Velocity

The relative velocity (also or ) is the velocity of an object or observer B in the rest frame of another object or observer A.
Velocity of B relative to A is =

V_{b} - V_{a}

definition

Relative Acceleration

The relative acceleration (also or ) is the acceleration of an object or observer B in the rest frame of another object or observer A.
Acceleration of B relative to A is

= a_{b} - a_{a}

example

Distance of closest approach

Example:
Two ships A and B are 10 km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. What is their distance of closest approach and how long do they to reach it?
Solution:
Ships A and B are moving with same speed 20 km/h in the directions shown in figure. It is two dimensional, two body problem with zero acceleration.
Let us find

{\vec{v}}_{B A}

{\vec{v}}_{B A} = {\vec{v}}_{B} - {\vec{v}}_{A}

Here,

| {\vec{v}}_{B A} | = \sqrt{{(20)}^{2} + {(20)}^{2}} = 20 \sqrt{2} k m / h

i.e.,

{\vec{v}}_{B A} i s 20 \sqrt{2}

km/h at an angle of

45^{\circ}

from east towards north. Thus, the given problem can be simplified as:
A is at rest and B is moving with

{\vec{v}}_{B A}

in the direction shown in figure.
Therefore, the minimum distance between the two is

s_{m i n} = A C = A B \sin 45^{\circ}

= 10 (\frac{1}{\sqrt{2}}) k m

= 5 \sqrt{2} k m

And the desired time is:

t = \frac{B C}{| {\vec{v}}_{B A} |} = \frac{5 \sqrt{2}}{20 \sqrt{2}} (B C = A C = 5 \sqrt{2} k m)

= \frac{1}{4} h

= 15 m i n

definition

Velocity of approach and separation

Velocity of approach or separation is defined as the rate of change of relative displacement between two bodies (i.e. how fast a body approaches another body). Velocity of approach is defined when the displacement between the bodies is decreasing and separation when the displacement between the bodies is increasing.
Example:
When two bodies approach each other with different uniform speeds, the distance between them decreases by

120 m

per every

1 m i n

. Then, the velocity of approach is

120 m / m i n = 2 m / s

example

Problems on relative motion between point objects

Bodies are dropped from a height in successive intervals of half a second. The relative velocity of one with respect to the other is:
At any tme,

v_{1} = u + g t_{1}

v_{2} = u + g t_{2}

t_{1} = t_{2} + 0.5

(Relative velocity)

= v_{1} - v_{2} = g (t_{1} - t_{2}) = g \times 0.5 = \frac{g}{2}

definition

Relative Motion between finite length objects

Example: A car is moving with velocity

3 \hat{i}

m/s find its velocity relative to a bike moving with

1.5 \hat{i}

m/s.

Solution:
Relative velocity of car with respect to bike:

3 \hat{i} - 1.5 \hat{i} = 1.5 \hat{i} m / s

example

Relative motion between point objects moving in two dimensions

Let the velocity of one object is:

\vec{v} = v_{1} \hat{i} + v_{2} \hat{j}

Velocity of the other object is:

\vec{u} = u_{1} \hat{i} + u_{2} \hat{j}

Relative velocity =

\vec{v} - \vec{u} = (v_{1} - u_{1}) \hat{i} + (v_{2} - u_{2}) \hat{j}

example

Triangle Law of Vector addition in Relative Motion

A swimmer can swim in still water with speed

v

and the river flowing with velocity

\frac{v}{2}

. To cross the river in shortest distance, he should swim making an angle

θ

with the upstream. Find ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance.
For shortest distance , Time taken

= \frac{W}{V_{b} s i n θ}

For shortest time , Time taken

= \frac{W}{V_{b}}

Ratio of times taken for shortest time to that of shortest path

= \frac{\frac{W}{V_{b}}}{\frac{W}{V_{b} s i n θ}} = s i n θ

example

Rain-Man Problem

A man runs with a velocity of

8 k m / h

towards east. The velocity of rainfall is

4 k m / h

vertically downwards. The angle at which the man should hold the umbrella so as to protect himself from the rain:

r =

rain

, g =

ground

& m =

manWe have,

\vec{v_{r m}} = \vec{v_{r g}} + \vec{v_{m g}}

Resultant

\vec{v_{r m}}

makes an angle of

θ = \tan^{- 1} \frac{8}{4} = {63.43}^{o}

with verticalSo, man should rotates his umbrella in anticlockwise direction such that its makes an angle of

{63.43}^{o}

with vertical.

example

Rain-Man Problem where rainfall is inclined to man's velocity

A man wearing a hat of extended length

12 c m

is running in rain falling vertically downwards with speed

10 m / s

. The maximum speed with which man can run, so that rain drops do not fall on his face (the length of his face below the extended part of the hat is

16 c m

)

V_{r a i n} = 10 m / s

(in negative

y

direction)
Let, velocity of man =

v

Also from the above diagram we get the value of

θ

t a n θ = \frac{16}{12} = \frac{4}{3}

then, velocity of rain w.r.t man

V_{r a i n / m a n}

v

(opposite to man i.e. in negative

x

direction)
For the required condition :

t a n θ = \frac{10}{v} = \frac{4}{3}

v = \frac{10 \times 3}{4} = 7.5 m / s

example

Objects moving upstream or downstream in a river

A boat takes

2 h o u r s

to travel

8 k m

and back in a still water lake. With water velocity of

4 k m / h

, the time taken for going upstream of

8 k m

and coming back is given by:

Speed of boat relative to water

= 8 k m / h

Speed downstream

= 12 k m / h

Speed upstream

= 4 k m / h

Time taken with water flow

= \frac{8}{12} + \frac{8}{4} h r s = 2 + \frac{2}{3} h r s = 160 m i n

example

Problem Analogous to Objects moving upstream or downstream in a river

Example:A man can row a boat with

4

km/h in still water. If he is crossing a river where the current is

2

km/h. Width of river is

4

km.How long will it take him to row

2

km up the stream and then back to his starting point?Solution:
Given :

V_{M W} = 4

km/hr

V_{R} = 2

km/hr
Distance covered in upstream or downward motion

d = 2

km each
For upstream motion of the boat :
Velocity of the man

V_{M} = V_{M W} - V_{W} = 4 - 2 = 2

km/h

∴

Time taken to go upstream

t_{1} = \frac{2}{2} = 1

hr

For downward motion of the boat or (return journey) :
Velocity of the man

V_{M} = V_{M W} + V_{W} = 4 + 2 = 6

km/h

∴

Time taken to go downstream

t_{2} = \frac{2}{6} = \frac{1}{3}

∴

Total time taken

T = t_{1} + t_{2} = 1 + \frac{1}{3} = \frac{4}{3}

hrs

example

Problems on objects crossing a river

A man who can swim at a speed

v

relative to the water wants to cross a river of width

d

, flowing with a speed

u

. The point opposite him across the river is A.
For reaching A, he must have a velocity component

0

along the flow of river. Hence, in the frame of river he must have a component

- u

along the river to have zero component in ground frame and hence perpendicular to the flow, he has velocity

\sqrt{v^{2} - u^{2}}

relative to both frames, Hence, the time taken

= \frac{d}{\sqrt{v^{2} - u^{2}}}

example

Object crossing a river where relative velocity is inclined to river velocity

At a harbour, a boat is standing and wind is blowing at a speed of

\sqrt{2}

m/sec. due to which, the flag on the boat flutters along northeast. Now the boat enters in to river, which is flowing with a velocity of

2 m / s

due north. The boat starts with zero velocity relative to the river and its constant acceleration relative to the river is

0.2 m / s e c^{2}

due east. The direction in which the flag will flutter at

10

seconds is given by:

V_{w} = 1 \hat{i} + 1 \hat{j}

V = a t

V = (0.2) 10 = 2 m / s e c .

V_{b o a t} = 2 \hat{i} + 2 \hat{j}

V_{w / b o a t} = V_{w} - V_{b o a t}

V_{w / b o a t} = (1 \hat{i} + 1 \hat{j}) - (2 \hat{i} + 2 \hat{j}) = - 1 \hat{i} - 1 \hat{j}

So, the flag will flutter towards south-west.

example

Problems on objects crossing a river to reach certain point on the opposite bank

A man can swim in still water with a velocity

5 m / s

. He wants to reach at directly opposite point on the other bank of a river which is flowing at a rate of

4 m / s

. River is

15 m

wide and the man can run with twice the velocity as compared with velocity of swimming. If he swims perpendicular to river flow and then run along the bank, then time taken by him to reach the opposite point will be given by:
Time taken in crossing the river

T_{1} = \frac{d}{V_{m r}}

T_{1} = \frac{15}{5}, T_{1} = 3 s e c

Distance travel in this time along the river

D = V_{R} \times T_{1}

D = 4 \times 3, D = 12

T_{2} = \frac{D}{2 V_{m r}}, T_{2} = 1.2 s e c

T = T_{1} + T_{2}, T = 4.2 s e c

example

Relative motion in river crossing like situations

River-crossing scenarios are when a boat moves from one bank to another to cross a river flowing with a given velocity. An analogous scenario is discussed below.
A man can walk at a speed of

2

m/s. He walks in north direction on a platform of width

12

m moving east at a speed of

3

m/s. Find the displacement of the man relative to the platform along the direction of the motion of platform.

v_{m} = 2 \hat{j}

v_{p} = 3 \hat{i}

v_{m, p} = 2 \hat{j} - 3 \hat{i}

s_{m, p} = 2 t \hat{j} - 3 t \hat{i}

But,

s_{m, p} = s_{p} \hat{i} + 12 \hat{j}

From above two equations,

s_{p} = - 18 m

example

Relative motion between an object in linear motion and a projectile

There is a toy plane moving horizontally at a height of 19.6m at a velocity of 50m/s. An object is dropped from the plane then the distance traveled by the plane at the instance object hits the ground has to be determined:

h = 1 / 2 g t^{2}

19.6 = 1 / 2 \times 9.8 \times t^{2}

t = 2 s

Distance moved by the plane in this interval of time is

50 \times 2 = 100 m

example

Relative Motion between freely falling body and a projectile

Due to failure of engine of a rocket it started falling freely after

2

seconds an astronaut takes a horizontal velocity

u = 5

m/s to come out of it. Astronaut lands at a horizontal distance of

50

m from the rocket. The height of jump of the astronaut from the rocket has to be determined:

d = u \times t

50 = 5 \times t

t = 10 s

Initial velocity of the rocket at the instant of jump =

10 \times 2

h = u t + 1 / 2 g t^{2}

h = 5 \times 10 + 1 / 2 \times 10 \times 10^{2}

h = 550 m

example

Problems on condition of collision between a projectile and an object in uniform motion

Two canons installed at the top of a cliff

10 m

high fire a shot each with speed

5 \sqrt{3} m s^{- 1}

at some interval. One canon fires at

60^{\circ}

with horizontal whereas the second fires horizontally, what are the coordinates of point of collision of shots?
Let

t_{1} & t_{2}

be the times of shot from canon at

and from horizontal canon to reach the point of collision.
For first canon:

y = - 5 \sqrt{3} \sin 60 t_{1} + \frac{1}{2} g t_{1}^{2} . . . (1) & x = 5 \sqrt{3} \cos 60 t_{1} . . . (2)

For second canon:

y = \frac{1}{2} g t_{2}^{2} . . . (3) & x = 5 \sqrt{3} t_{2} . . . (4)

Equating:

(2) & (4) t_{1} = 2 t_{2} . . . (5)

Equating:

(1) & (3) - 5 \sqrt{3} \sin 60 t_{1} + \frac{1}{2} 10 t_{1}^{2} = \frac{1}{2} 10 t_{2}^{2} . . . (6)

From

(5) & (6), t_{1} = 2 s, t_{2} = 1 s

so,

x = 5 \sqrt{3} m & y = 5 m

example

Problems on relative motion between two projectile projected

A cannon fires successively two shells with velocity

v_{0} = 250 m / s

; the first at the angle

θ_{1} = 60^{\circ}

and the second at the angle

θ_{2} = 45^{\circ}

to the horizontal, the azimuth being the same.
Neglecting the air drag, the time interval(in seconds) between firings leading to the collision of the shells: (Take

g = 10 m / s^{2}

and Round-off your answer to the nearest integer.)
Let the shells collide at the point

P (x, y)

. If the first shell takes

t s

to collide with second and

Δ t

be the time interval between the firings, then

x = v_{0} \cos θ_{1} t = v_{0} \cos θ_{2} (t - Δ t)

................(1)
and

y = v_{0} \sin θ_{1} t - \frac{1}{2} g t^{2}

= v_{0} \sin θ_{2} (t - Δ t) - \frac{1}{2} g {(t - Δ t)}^{2}

.............. (2)
From equation (1)

t = \frac{Δ t \cos θ_{2}}{\cos θ_{2} - \cos θ_{1}}

...................................(3)
From equations (2) and (3)

Δ t = \frac{2 v_{0} \sin (θ_{1} - θ_{2})}{g (\cos θ_{2} + \cos θ_{1})}

Δ t \neq 0

example

Relative motion in projectiles

Example:
CE and DF are two walls of equal height (

20

meter) from which two particles A and B of same mass are projected as shown in the figure.A is projected horizontally towards left while B is projected at an angle

37^{0}

(with horizontal towards left) with velocity 1

5 m / s

. If A always sees B to be moving perpendicular to EF, then find the range of A on ground.
Solution:From given conditions :

V_{A} = V_{B} c o s 37^{0} = 15. \frac{4}{5} = 12 m / s .

∴

time of flight of A (t)

= \sqrt{\frac{2 \times 20}{10}} =

2 s.

\Rightarrow R a n g e = V_{A} t = 24 m

definition

Distance-time graph

A distance-time graph is a graph of distance v/s time. It lies in first quadrant only as distance is always positive. Also, it is increasing in nature. The attached plot shows a distance-time graph.

diagram

Displacement time graph

Displacement-time graph is a plot between displacement and time. Note that displacement can be both positive and negative. Also since it is a vector, displacement-time graph is drawn for one-dimension of motion only. The attached figure shows a plot of displacement-time graph.

result

Velocity from displacement-time graph

Velocity of a particle is equal to the slope of its displacement-time graph.

v = \frac{s_{2} - s_{1}}{t_{2} - t_{1}}

In the given graph,

v = \frac{40 - 20}{4 - 2} = 10 m / s

Velocity may be positive, negative or zero since it is a vector quantity.
Note: When we draw graph of a vector (say displacement), each component (like component in x-direction) of the vector is drawn on a separate graph. So, for a 2-D motion, two graphs are needed to completely describe the motion of a particle.

definition

Average velocity from a displacement-time graph

Average velocity equals the ratio of net displacement and the time taken.
Example:
In the given plot, for

t \in (0, 10)

,
Displacement

s = - 4 - 0 = - 4

Hence, average velocity is given by

v = s / t = - 4 / 10 = - 0.4 m / s

diagram

Displacement time graph for rest, uniform motion and uniform acceleration

Displacement time graph for rest is a straight line with zero slope.
Displacement time graph for uniform motion is a straight line with non-zero slope.
Displacement time graph for uniform acceleration is a parabola.
The three scenarios are shown in the attached plot.

definition

Velocity-time graph

Velocity-time graph is a plot between velocity and time. Magnitude of velocity at a given instant is equal to its instantaneous speed. It is drawn for 1-D motion only and can take both positive and negative values. Attached figure shows a velocity-time graph.

result

Acceleration from velocity-time graph

Acceleration of a particle is equal to the slope of a velocity-time graph.

a = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}

In the given graph,

a = \frac{40 - 20}{4 - 2} = 10 m / s^{2}

Acceleration may be positive, negative or zero as it is a vector quantity.

result

Displacement from velocity-time graph

Displacement of a particle in a given time-interval is equal to the total area under the velocity-time graph in the given time-interval.
In the given graph, displacement is found as area of trapezium ABCD.

s = \frac{1}{2} \times D C \times (A D + B C)

Note: Since, displacement is a vector quantity, Area below the time-axis is considered as negative and above it is considered as positive.

diagram

Velocity Time graph for rest, uniform motion and uniform acceleration

result

Acceleration time graphs

Acceleration time graph is plotted between acceleration and time.
(a) A constant non-zero acceleration time graph indicates a body under constant acceleration, example: free-falling body
(b) A constant zero acceleration-time graph indicates a body under uniform motion example: body sliding on a frictionless surface.
(c) A non-constant acceleration time graph indicates varying acceleration. For these, equations of motions are invalid.

Area under the acceleration-time graph in a time-interval is equal to the net change of velocity in the time-interval.

definition

Area under acceleration time graph

Change in velocity of a particle can be evaluated as area under the acceleration-time graph.

as

a = \frac{d v}{d t}

d v = a d t

V e l o c i t y = \int a d t

(Which is area under the curve.)

definition

Distance-time graph from displacement-time graph for 1-D motion

In 1-D motion, distance is equal to the sum of absolute displacements of all infinitesimally small time-intervals.

result

Distance in nth second(distance-time graph)

Distance traveled in nth second can be found using distance time-graph. In the given figure, distance after

21 s

d_{2} = 21 m

and after

20 s

d_{1} = 20 m

. Hence, distance traveled in

21^{st}

sec is

d_{2} - d_{1} = 1 m

diagram

Displacement time graph for motion described piece-wise

Attached figure shows a motion described piece-wise.
For

0 < t < 5

, motion is uniform and velocity is positive, hence displacement is increasing.
For

5 < t < 10

, motion is uniform and velocity is negative, hence displacement is decreasing.
For

10 < t < 15

, body is at rest and velocity is zero, hence displacement is constant.

result

Average velocity from velocity-time graph

Instantaneous velocity at a given time-instant for a particle can be found from the y-intercept of its velocity-time graph.
From a particle's velocity-time graph, its average velocity can be found by calculating the total area under the graph and then dividing it by the corresponding time-interval.
For a particle with uniform acceleration, velocity-time graph is a straight line. Its average velocity is given by

v_{a v g} = (v_{i} + v_{f}) / 2

. From the graph, this can be found by drawing the y-intercepts of initial and final velocities and then drawing the mid-point.
In the given graph, instantaneous velocity

= v_{1}

t = t_{1}

and

v_{2}

t = t_{2}

.
Average velocity is found at E, i.e. mid-point of C and D.

definition

Velocity time graph for motion described piece-wise

Example:
Find the total displacement for the given graph in

0 < t < 7

.
Solution:
For

0 < t < 5

, displacement is given by

s_{1} = \frac{1}{2} \times 5 \times 50 = 125 m

For

5 < t < 7

, displacement is given by

s_{2} = 2 \times 50 = 100 m

Total displacement,

s = s_{1} + s_{2} = 225 m

definition

Interpretation of acceleration-time graph

Acceleration vs. time graphs tell us about an object's velocity in the same way that velocity vs. time graphs tell us about an object's displacement. The change in velocity in a given time interval is equal to the area under the graph during that same time interval.

definition

Acceleration Vs Time Graph for piece-wise execution of motion

Acceleration-time graph for a particle describing motion piece-wise for separate time intervals.

example

Inferring motion plots from given plots

Following the relationship between various motion parameters plots can be inferred for other parameter.

example

Interpreting Relative Motion Through Graphs

Relative position between the particles is not changing over the passage of time therefore relative position has become constant.

example

Velocity-displacement graph

Example:
The velocity

(v)

- distance

(s)

graph for an airplane travelling on a straight runway is shown. Find the acceleration (in

m / s^{2}

) of the plane at

s = 50 m

.
Solution:

a = v \frac{d v}{d x} = 20 \times \frac{40}{100} = 8

definition

Feasibility of a graph

A graph is feasible if it does not violate the definition of a quantity. Some guidelines to be followed to check feasibility of a graph are:
1. Distance and speed are greater than zero.
2. Distance-time plot is increasing in nature.
3. Displacement/distance of a graph cannot change suddenly.
4. Quantity cannot take two values at an instant of time.
5. Graph does not exist in third or fourth quadrant since time is taken positive in normal scenarios.

Monday, July 31, 2017

Kinematics2

Projectile clearing multiple obstacles

Projectile on an inclined Plane

Relative Displacement

Relative Velocity

Relative Acceleration

Distance of closest approach

Velocity of approach and separation

Problems on relative motion between point objects

Relative Motion between finite length objects

Relative motion between point objects moving in two dimensions

Triangle Law of Vector addition in Relative Motion

Rain-Man Problem

Rain-Man Problem where rainfall is inclined to man's velocity

Objects moving upstream or downstream in a river

Problem Analogous to Objects moving upstream or downstream in a river

Problems on objects crossing a river

Object crossing a river where relative velocity is inclined to river velocity

Problems on objects crossing a river to reach certain point on the opposite bank

Relative motion in river crossing like situations

Relative motion between an object in linear motion and a projectile

Relative Motion between freely falling body and a projectile

Problems on condition of collision between a projectile and an object in uniform motion

Problems on relative motion between two projectile projected

Relative motion in projectiles

Distance-time graph

Displacement time graph

Velocity from displacement-time graph

Average velocity from a displacement-time graph

Displacement time graph for rest, uniform motion and uniform acceleration

Velocity-time graph

Acceleration from velocity-time graph

Displacement from velocity-time graph

Velocity Time graph for rest, uniform motion and uniform acceleration

Acceleration time graphs

Area under acceleration time graph

Distance-time graph from displacement-time graph for 1-D motion

Distance in nth second(distance-time graph)

Displacement time graph for motion described piece-wise

Average velocity from velocity-time graph

Velocity time graph for motion described piece-wise

Interpretation of acceleration-time graph

Acceleration Vs Time Graph for piece-wise execution of motion

Inferring motion plots from given plots

Interpreting Relative Motion Through Graphs

Velocity-displacement graph

Feasibility of a graph

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