Physics chapter name Motion in a Straight line.

Motion in a
3
BASIC DEFINITIONS
Straight Line
3. Displacement of an object is independent of the path
Mechanics : Branch of physics, which deals with the study of followed by the object but distance depends upon path.
objects in rest and in motion. 4. Displacement and distance both have same unit as that of
Statics : Study of objects at rest or in equilibrium. length i.e. metre.
Kinematics : Study of motion of objects without considering the
cause of motion. Distance
5. ³1
Dynamics : Study of motion of objects considering the cause of |Displacement|
motion.
Rest : An object is said to be at rest if it does not change its 6. For a moving body distance always increases with time
position with time, with respect to its surrounding (a reference 7. For a body undergoing one dimensional motion, in the same
point which is generally taken as origin in numerical problems) direction distance = | displacement |. For all other motion
Motion : An object is said to be in motion if it changes its position distance > | displacement |.
with time, with respect to its surroundings.
Rest and motion are relative terms. SPEED
Point mass/Point object : An object is said to be a point mass if It is the distance travelled per unit time by an object. It is a scalar
during its motion it covers distance much greater than its own quantity. It cannot be negative.
size. Uniform speed : An object is said to be moving with a uniform
One dimensional motion : An object travels in a straight line. It is speed, if it covers equal distances in equal intervals of time,
also called rectilinear or linear motion. The position change of howsoever small the time intervals may be.
the object with time in one dimension can be described by only Non-uniform speed : If an object covers unequal distances in
one co-ordinate. equal interval of time or equal distances in unequal interval of
Ex. A stone falling freely under gravity. time.
Two dimensional motion or motion in a plane : For an object Instantaneous speed : The speed of an object at a particular instant
travelling in a plane two coordinates say X and Y are required to of time is called the instantaneous speed.
describe its motion.
Ex. An insect crawling over the floor. Ds ds
Instantaneous speed, Vinst = lim = .
Three dimensional motion : An object travels in space.To describe Dt dt
Dt ® 0
motion of objects in three dimension require all three coordinates Average speed : It is ratio of the total distance travelled by the
x, y and z. object to the total time taken.
Ex. A kite flying in the sky.
Dx
DISTANCE AND DISPLACEMENT Average speed Vav =
Dt
Distance or Path length : The length of the actual path travelled Dimensions : [M0LT-1]; Unit: In SI systems.
by an object during motion in a given interval of time is called
the distance travelled by that object or path length. It is a scalar VELOCITY
quantity. It is the displacement of an object per unit time. It is a vector
Displacement : It is the shortest distance between the initial and quantity. It can be positive negative or zero.
final position of an object and is directed from the initial position Uniform velocity : An object is said to be moving with a uniform
to the final position. It is a vector quantity. velocity, if it covers equal displacements in equal intervals of
time, howsoever small the time intervals may be.
Keep in Memory Non-uniform velocity : If an object covers unequal displacements
1. Displacement may be positive, negative or zero but distance in equal interval of time or equal displacements in unequal interval
is always positive. of time.
Instantaneous velocity : The velocity of an object at a particular
2. Displacement is not affected by the shift of the coordinate
axes. instant of time is called the instantaneous velocity.

, 36 PHYSICS

r r r Variable or non-uniform acceleration : If the velocity of body
Dr dr
Instantaneous velocity, Vinst = lim = . changes in different amounts during same time interval, then the
Dt ® 0 Dt dt
acceleration of the body is known as variable acceleration.
Average Velocity : It is ratio of the total displacement to the total
Acceleration is variable if either its direction or magnitude or
time taken.
r both changes with respect to time. A good example of variable
Dr acceleration is the acceleration in uniform circular motion.
Average velocity, vav =
Dt
EQUATIONS FOR UNIFORMLY ACCELERATED MOTION
Dimensions : [M0LT–1] ; Unit: In SI system, m/s
When the motion is uniformly accelerated i.e., when acceleration
Keep in Memory is constant in magnitude and direction :
(i) v = u + at
1. | Average velocity | can be zero but average speed cannot be
zero for a moving object. 1 2
(ii) s = ut + at
2
|Average velocity|
2. £1 (iii) v2 – u2 = 2as
Average speed
where u = initial velocity; v = final velocity;
3. | Instantaneous velocity | = Instantaneous speed. a = uniform acceleration and
4. A particle may have constant speed but variable velocity. It
s = distance travelled in time t,
happens when particle travels in curvilinear path.
5. If the body covers first half distance with speed v1 and æ u + vö
(iv) s =ç t
next half with speed v2 then è 2 ÷ø
2v1v2 a
Average speed v = (v) Distance travelled in nth second sn = u + (2n - 1) ;
v1 + v2 2
6. If a body covers first one-third distance at a speed v1, next sn = distance covered in n th second
one-third at speed v2 and last one-third at speed v3, then Above equations in vector form
3v1 v2 v3 1
Average speed v = v = u + a t, s = u t + a t 2 ,
v1 v 2 + v 2 v3 + v1 v3 2
7. If a body travels with uniform speed v1 for time t1 and with r2 r2 rr rr rr rr
v - u = 2a.s (or v.v + u.u = 2a.s )
uniform speed v2 for time t2, then
r 1 r r
v t +v t s = ( u + v) t
Average speed v = 1 1 2 2 2
t1 + t2 When displacement (s) is given as a function of time t [s = f(t)]
then
ACCELERATION
The rate of change of velocity with respect to time is called s µ t0 Body at rest v= 0 a= 0
acceleration. It is a vector quantity.
r s µ t1 Uniform velocity;
Let velocity changes by Dv during some interval of time Dt .
r acceleration zero v µ t0 a= 0
Average acceleration a av is given by
r
r Dv s µ t2 Uniform acceleration v µ t1 a µ t0
aav =
Dt
r s µ t 3 or more Non uniform
Instantaneous acceleration a is given by
r r acceleration v µ t 2 or more a µ t1 or more
r Dv dv
a = lim = We use calculus method (integration and differentiation)
Dt ®0 Dt dt
for displacement, velocity, acceleration as a function of time.
Its SI unit is meter/sec2 (ms–2).
ds dv
A body moving with uniform velocity has zero acceleration. It We know that v =
dt ò
Þ s = v dt ; a =
dt ò
Þ v = a dt ;
means that neither its speed nor its direction of motion is changing
with time. when a = f(s)
dv
Uniform acceleration : If the velocity of the body changes in
equal amount during same time interval, then the acceleration of
a=v
ds ò
Þ a ds = v dv , ò
the body is said to be uniform. Acceleration is uniform when where s = displacement, v = instantaneous velocity,
neither its direction nor magnitude changes with respect to time. a = instantaneous acceleration

, Motion in a Straight Line 37

VERTICAL MOTION UNDER GRAVITY
1 2 g
(i) For a body thrown downward with initial velocity u from a h= gt ; h th = (2n - 1)
2 n 2
height h, the equations of motion are
(iii) When a body is thrown upwards with initial velocity u, the
v = u +gt ; v 2 = u 2 + 2gh equations of motion are
v = u – gt
1 2 g
h = ut + gt ; h th = u + (2n - 1)
2 n 2 1 2
h = ut – gt
(ii) If initial velocity is zero, then the equations are 2
v2 = u2 – 2gh.
v = gt ; v = 2gh

To summarise :
Motion of a body



Velocity constant Accelerated motion
Displacement
Velocity =
Time
Uniformly accelerated Non-uniformly
motion accelerated motion
v = u + at ds dv dv
v= a= =v
1 dt dt ds
s = ut + at 2
2 Differentiation Differentiation
2 2 Displacement Velocity Acceleration
v - u = 2as
a Integration Integration
s nth = u + (2n - 1)
2
1
s = (u + v) ´ t
ò ds = ò v dt ò dv = ò a dt
2

N OTE: Calculus method as shown in non-uniformly accelerated Keep in Memory

motion may also be used for uniformly accelerated motion. 1. The direction of average acceleration vector is the direction
r r
Uniformly Accelerated Motion : A Discussion r vf - vi
While using equations of motion we can have two approaches. of the change in velocity vector. a =
t
Approach 1 : Take a = +ve when velocity increases and r r r r r
a has a direction of v f - v i = v f + (– v i )
a = –ve when velocity decreases. r r
Take rest of physical quantities such as u, v, t and s as positive. i.e., the resultant of vf and – v i
Approach 2 : (Vector method)
2. There is no definite relationship between velocity vector
Assume one direction to be positive and other negative. Assign
and acceleration vector.
sign to all the vectors (u, v, a, s), +ve sign is given to a vector
which is directed to the positive direction and vice-versa 3. For a body starting from rest and moving with uniform
– + acceleration, the ratio of distances covered in t 1 sec.,
+




t2 sec, t3 sec, etc. are in the ratio t 12 : t22 : t32 etc.
+




4. A body moving with a velocity v is stopped by application
– of brakes after covering a distance s. If the same body moves
–
with a velocity nv, it stops after covering a distance n 2s by
Normally the direction taken is as drawn above. But it is important the application of same retardation.
to note that you can take any direction of your choice to be positive
Example 1.
and the opposite direction to be negative. The displacement of particle is zero at t = 0 and
displacement is ‘x’ at t = t. It starts moving in the positive
N OTE: The second method (or approach) is useful only when
x-direction with a velocity which varies as v = k x where
there is reversal of motion during the activity concerned. k is constant. Show that the velocity varies with time.

, 38 PHYSICS

Solution : r r
r v - vi 5jˆ - 5iˆ r 1 1
v = k x or
dx
= k x or
dx
= k dt
a= f = Þ a = – $i + $j
Dt 10 2 2
dt x
Given that when t = 0, x = 0 and when t = t, x = x, r FG 1 IJ + FG 1 IJ
2 2 1
m/s2.
x
dx
t |a |=
H 2K H 2 K =
2
Hence ò x
ò
= k dt ;
Keep in Memory
0 0
x
é - 1 +1 ù 1. An object moving under the influence of earth's gravity in
x t êx 2 ú which air resistance and small changes in g are neglected is
ú = k [ t ]0
t
\ ò x -½ dx = k ò dt or ê called a freely falling body.
1
ê - + 1ú
0 0 2. In the absence of air resistance, the velocity of projection
ëê 2 ûú 0
is equal to the velocity with which the body strikes the
kt ground.
or 2 x = k t or x=
2 3. Distance travelled by a freely falling body in 1st second is
2 always half of the numerical value of g or 4.9 m, irrespective
ék tù k t
Now, v = k ´ ê ú = of height h.
ë2û 2 4. For a freely falling body with initial velocity zero
Thus the velocity varies with time. (i) Velocity µ time (v = gt)
Example 2. (ii) Velocity µ Distance fallen (v2 = 2gs)
A particle covers each 1/3 of the total distance with speed
vl, v2 and v3 respectively. Find the average speed of the æ 1 ö
(iii) Distance fallen a (time)2 ç s = gt 2 ÷ , where g is the
particle. è 2 ø
Solution : acceleration due to gravity.
Average speed 5. If maximum height attained by a body projected vertically
Total distance travelled s upwards is equal to the magnitude of velocity of projection,
v= = then velocity of projection is 2g ms–1 and time of flight is 4
Total time taken s s s
+ + sec.
3v1 3v 2 3v 3
6. If maximum height attained by a body projected upward is
3v1 v 2 v 3 equal to magnitude of acceleration due to gravity i.e., 'g',
Þ v=
v1v 2 + v 2 v 3 + v 3 v1 the time of ascent is 2 sec. and velocity of projection is
Example 3.
A cheetah can accelerate from 0 to 96 km/h in 2 sec., whereas g 2.
a cat requires 6 sec. Compute the average accelerations for 7. Ratio of maximum heights reached by different bodies
the cheetah and cat. projected with velocities u1, u2, u3 etc. are in the ratio of
Solution : u12 : u 22 : u 33 etc. and ratio of times of ascent are in ratio of
r r
r | vf - vi | 96km / h - 0 u1 : u2 : u3 etc.
For cheetah | a av| = =
Dt 2sec 8. During free fall velocity increases by equal amount every
1000 m decend and distance covered during 1st, 2nd, 3rd seconds
96 ´ of fall, are 4.9m, 14.7m, 24.5m.
3600 sec
= = 15 m/s2
2 sec u=0 t=0
10 4.9 m
96 ´ 9.8 m/s t=1
r
For cat | a av| = 36 = 5 m/s2. 14.7 m
6 19.6 m/s t=2
Example 4.
A particle is moving in east direction with speed 5 m/s.
After 10 sec it starts moving in north direction with same 24.5 m
speed. Find average acceleration.
Solution : N 29.4 m/s t=3
r r
| v f| = | v i| = 5 m/s
9. If a body is projected horizontally from top of a tower, the
Acceleration ¹ 0
W E time taken by it to reach the ground does not depend on
(due to change in direction the velocity of projection, but depends on the height of
of velocity Av. acceleration,
2h
r
D v = vr f – vr i = vr f + (– vr i) S
tower and is equal to t =
g
.