Contents
Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
What are the 3 Derivations of Motion?
When a body moves without changing its direction, the motion is naturally along a straight line. It is also known as rectilinear motion.
Vertical fall of a body under gravity, motion of a car on a straight road, etc. are examples of such motion. The discussions henceforth in this chapter will be restricted to rectilinear motions only.
Representation of the physical quantities of motion: For the rectilinear motion of a particle, the straight line of motion itself may be chosen as an axis (say, the x-axis) and a point O on it as the origin [Fig.].
Obviously, the motion is one-dimensional.
There is a basic difference in the representations of the scalar and the vector quantities related to rectilinear motion. The scalar quantities like distance travelled and speed have magnitudes only, and are always expressed by positive numbers. However, two directions, exactly opposite to each other, exist for the motion in a straight line. So a vector quantity is expressed by a positive number for one direction and by a negative number for the exactly opposite direction.
In Fig., the direction towards right may be taken as positive; then the direction towards left naturally becomes negative. As an example, we may consider the vertical motion of a particle under gravity. For a downward motion, we may take each of displacement, velocity and acceleration to be positive. Then, for an upward motion, displacement and velocity would be negative; but acceleration would still remain positive since the acceleration due to gravity (g) is always directed downwards.
It is important to note that simple algebraic operations are sufficient for calculations involving any quantit a scalar or a vector, in a rectilinear motion. It means that, vector algebra is not at all necessary even for the calculations involving vector quantities of a rectilinear motion.
For motion along a straight line,
- distance travelled by a particle = magnitude of its displacement,
- speed of the particle at any point = magnitude of its velocity at that point
- displacement, velocity and acceleration vectors are along the same straight line.
Rectilinear Motion with Uniform Velocity
If a particle moves with a uniform velocity, its acceleration is zero. Let v be the uniform velocity of a particle and s be its displacement in time t. Therefore, according to the definition of uniform velocity, the particle moves a distance v × 1 in 1 s, v × 2 in 2 s, etc.
∴ In t s its displacement is v × t.
∴ s = vt ……. (1)
i.e., displacement = uniform velocity × time.
Numerical Examples
Example 1.
A person travels half of a distance at an average velocity of 24 km ᐧ h-1. At what average velocity should he move to cover the second half of the path so that his average velocity for the total path becomes 32 km ᐧ h-1?
Solution:
Let the total length of the path 2s km.
∴ Time required to cover the first half of the path = \(\frac{s}{24}\)h.
If the man travels the total path with an average velocity of 32 km ᐧ h-1, then the total time taken by him = \(\frac{2 s}{32}\) = \(\frac{s}{16}\)h.
∴ Time required to cover the second half of the path
= \(\frac{s}{16}\) – \(\frac{s}{14}\) = \(\frac{s}{48}\)h
∴ Average velocity of second half
\(=\frac{\text { distance }}{\text { time }}\) = \(\frac{s}{\frac{s}{48}}\) = 48 km ᐧ h-1
Rectilinear Motion with Uniform Acceleration
For a particle in motion, let
u = initial velocity
v = final velocity after a time t and corresponding uniform acceleration = a. Let the displacement = s in that time.
Then the above variables obey the following equations:
- v = u + at,
- s = ut + \(\frac{1}{2}\)at2
- v2 = u2 + 2as
- st = u + \(\frac{1}{2}\)a(2t – 1) [where st = the displacement in the t th second]
Derivations:
i) v = u + at
Let the initial velocity of a particle be u and its final velocity after time t be v.
∴ In time t, change in velocity = v – u
∴ Rate of change of velocity with time = \(\frac{v-u}{t}\) = a, by definition.
Hence, at = v – u or, v = u + at …….. (1)
For a particle starting from rest, u = 0 and thus,
v = at ….. (2)
In case of retardation, the relationship becomes
v = u – at
s = ut + \(\frac{1}{2}\)at2 …… (3)
ii) s = ut + \(\frac{1}{2}\)at2
If a particle with initial velocity u and uniform acceleration a attains final velocity y after time t the average velocity, considering the two end points, is \(\frac{u+v}{2}\).
The acceleration a implies that the velocity increases by a after every second. Hence, the velocity 1 s after the start of motion = u + a and 1 s before the end of motion = v – a
∴ Average velocity = \(\frac{u+a+v-a}{2}\) = \(\frac{u+v}{2}\)
Hence, the particle can be considered to have travelled a distance s with the average velocity \(\frac{(u+v)}{2}\) in time t.
Hence, displacement,
s = \(\frac{u+v}{2}\) × t = \(\frac{u+(u+a t)}{2}\) × t [∵ v = u + at]
or, s = \(\frac{2 u t}{2}\) + \(\frac{1}{2}\)at2 or, s = ut + \(\frac{1}{2}\)at2 ….. (4)
For a particle starting from rest, u = 0
So, s = \(\frac{1}{2}\)at2 ….. (5)
For a retarding particle,
s = ut – \(\frac{1}{2}\)at2 ….. (6)
iii) v2 = u2 + 2as
From equation (1) we have,
v = u + at
or, v2 = (u + at)2 = u2 + 2uat + a2t2
= u2 + 2a (ut + \(\frac{1}{2}\)at2)
or, v2 = u2 + 2as [using equation (4)] …… (7)
Hence, for a particle starting from rest
v2 = 2as ….. (8)
and in case of retardation
v2 = u2 – 2as
iv) st = u + \(\frac{1}{2}\)a(2t – 1)
Displacement in t seconds, s = ut + \(\frac{1}{2}\)at2, from equation (4).
Displacement in (t – 1) seconds,
s’ = u(t – 1) + \(\frac{1}{2}\)a(t – 1)2
Hence, the displacement in the t th second,
st = s – s’
or, st = ut + \(\frac{1}{2}\)at2 – {u(t – 1) + \(\frac{1}{2}\)(t – 1)2a}
= ut + \(\frac{1}{2}\)at2 – ut + u – \(\frac{1}{2}\)at2 + \(\frac{1}{2}\).2ta – \(\frac{a}{2}\)
= u + at – \(\frac{a}{2}\) = u + \(\frac{1}{2}\)a(2t – 1) ….. (10)
For a particle starting from rest
st = \(\frac{1}{2}\)a(2t – 1)
and in case of retardation
st = \(\frac{1}{2}\)a(2t – 1) ……. (11)
and in case of retardation
st = u – \(\frac{1}{2}\)a(2t – 1) …. (12)
Numerical Examples
Example 1.
A velocity of 60 km ᐧ h-1 of a train is reduced by the application of brakes. A retardation of 40 cm ᐧ s-2 is produced. After how much time will the train stop? What will be the velocity of the train after 20 s ?
Solution:
Given, u = 60 km ᐧ h-1 = \(\frac{60 \times 1000}{60 \times 60}\) = \(\frac{50}{3}\) m ᐧ s-1
a = 40 cm ᐧ s-2 = 0.4 m ᐧ s-2 and v = 0
Hence, from the relation v = u – at, we get
0 = \(\frac{50}{3}\) – 0.4 × t or, t = \(\frac{50}{3 \times 0.4}\) = 41.7
∴ The train will stop after 41.7 s.
The velocity after 20 s,
v = \(\frac{50}{3}\) – 0.4 × 20
= \(\frac{26}{3}\) m ᐧ s-1 = 8.7 m ᐧ s-1
= \(\frac{8.7 \times 60 \times 60}{1000}\) km ᐧ h-1
= 31.3 km ᐧ h-1
Example 2.
A body covers 200 cm in the first 2 s of motion and 220 cm in the next 4 s. Calculate the velocity 7 s after the start.
Solution:
We know, s = ut + \(\frac{1}{2}\)at2
∴ Putting t = 2 s
200 = u × 2 + \(\frac{1}{2}\)a × 4 = 2u + 2a
or, u + a = 100 ….. (1)
The displacement in (4 + 2) or 6 s = 200 + 220 = 420 cm
∴ 420 = u × 6 + \(\frac{1}{2}\)a × 36 = 6u + 18a
or, u+ 3a = 70 …. (2)
From equations (1) and (2) we get,
a = -15 cm ᐧ s-2 and u = 115 cm ᐧ s-1
Velocity after 7 s of motion,
v = u + at = 115 – 15 × 7 = 115 – 105 = 10 cm ᐧ s-1
Example 3.
A man is 9 m behind a train at rest. The train starts with an acceleration of 2 m ᐧ s-2 and simultaneously the man starts running. He is able to board the train somehow after 3 s. Find the acceleration of the man.
Solution:
While boarding the train, the positions of the man and of the train must be the same.
Let the acceleration of the man be a and the distance traversed by the train in 3 s be x.
We know, s = ut + \(\frac{1}{2}\)at2. Here u = 0 as the train as well as the man starts from rest.
So, for the train, x = \(\frac{1}{2}\) × 2 × (3)2 or, x = 9 m
Thus, the distance traversed by the man in this time = 9 + 9 = 18 m
Then, s = 18 m, u = 0 and t = 3 s
∴ a = \(\frac{2 s}{t^2}\) = \(\frac{2}{3^2}\) × 18 = 4 m ᐧ s-2
Example 4.
A particle moves with a uniform acceleration along a straight line. It covers 41 cm and 49 cm in the 6th and the 10 th seconds respectively. What will be the distance covered by the particle in 15 s?
Solution:
We know, distance covered in the nth second,
sn = u + \(\frac{1}{2}\)a(2n – 1)
According to the question,
s6 = u + \(\frac{1}{2}\)a(2 × 6 – 1) or, 41 = u + \(\frac{11}{2}\)a ….. (1)
and s10 = u + \(\frac{1}{2}\)a(2 × 10 – 1) or, 49 = u + \(\frac{19}{2}\)a ….. (2)
By solving equations (1) and (2) we get,
u = 30 and a = 2
Now, putting t = 15 in s = ut+ \(\frac{1}{2}\)at2 we get,
s = 30 × 15 + \(\frac{1}{2}\) × 2 × (15)2 = 450 + 225 = 675 cm
∴ The particle traverses 675 cm in 15 s.
Example 5.
A train begins its journey from station A and stops at station B after 45 min. C is a certain point between A and B where the train attains its maximum velocity of 50 km ᐧ h-1. If the train travels from A to C with a uniform acceleration and from C to B with a uniform retardation, calculate the distance between A and B.
Solution:
Let the train start from A with a uniform acceleration a1 and reach C in time t1. From here it travels to B in time t2 with a uniform retardation a2 [Fig.].
The maximum velocity of the train is v at the point C.
Let AC = s1 and CB = s2
So for the motion from A to C,
v = a1t1 and s1 = \(\frac{1}{2} a_1 t_1^2\)
and for the motion from C to B,
0 = v – a2t2 or, v = a2t2
So the distance between A and B,
Example 6.
A train moving with a constant acceleration crosses an observer standing on the platform. The first and the second compartments, each 15 m long, cross the observer in 2 s and 2.5 s, respectively. Find the velocity of the train when its first compartment just crosses the observer and also find its acceleration.
Solution:
Let the velocity and the acceleration of the train as its 1st compartment just reaches the observer be u and a, respectively.
Hence, displacement in 2 s
= length of the 1st compartment = 15 m
and displacement in (2 + 2.5) or, 4.5 s = total length of the two compartments = 2 × 15 = 30 m
Now from the equation s = ut+ \(\frac{1}{2}\)at2 we get,
∴ 15 = 2u + \(\frac{1}{2}\)a(2)2
or, u + a = \(\frac{15}{2}\) ….. (1)
and 30 = 4.5u + \(\frac{1}{2}\)a(4.5)2
or, 36u + 81a = 240
Solving equations (1) and (2), we get,
u = \(\frac{49}{6}\) and a = –\(\frac{2}{3}\)
The 1st compartment crosses the observer in 2 s; the velocity at that moment,
v = u + at = \(\frac{49}{6}\) – \(\frac{2}{3}\) × 2 = \(\frac{41}{6}\)
Therefore, the velocity and acceleration of the train as its 1st compartment just crosses the observer are \(\frac{41}{6}\) m ᐧ s-1 and –\(\frac{2}{3}\) ᐧ m ᐧ s-2 respectively.
Example 7.
A bullet with an initial velocity u penetrates a target. After penetrating a distance x, its velocity decreases by \(\frac{\boldsymbol{u}}{\boldsymbol{n}}\). How much farther will the bullet move through the target before it comes to rest?
Solution:
Let us assume that the retardation of the bullet inside the target is a and it is uniform.
After penetrating a distance x, the velocity,
v = u – \(\frac{u}{n}\) = \(\frac{u(n-1)}{n}\)
Example 8.
Starting from rest, a train travels a certain distance with a uniform acceleration α. Then it travels with a uniform retardation β and finally comes to rest again. If the total time of motioñ is t, find
(i) the maximum velocity attained and
(ii) the total distance travelled by the train.
Solution:
i) Let t1 be the time taken to travel a distance s1 with an acceleration α, and t2 be the time taken to travel a farther distance s2 with retardation β. Let the maximum velocity attained by the train be v.
Here t = t1 + t2
∴ For the motion of the train with acceleration α,
v = αt1 or, t1 = \(\frac{\nu}{\alpha}\) …. (1)
and v2 = 2αs1 or, s1 = \(\frac{v^2}{2 \alpha}\) …… (2)
Similarly, for the motion of the train with retardation β,
0 = v – βt2 or, t2 = \(\frac{\nu}{\beta}\) ….. (3)
and 02 = v2 – 2βs2 or, s2 = \(\frac{v^2}{2 \beta}\) ….. (4)
From (1) and (3) we get,
t = t1 + t2 = v\(\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) or v = \(\frac{\alpha \beta t}{\alpha+\beta}\)
ii) From (2) and (4), the total distance travelled,
s = s1 + s2 = \(\frac{v^2}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
= \(\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^2\) × \(\frac{1}{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right)\) = \(\frac{\alpha \beta t^2}{2(\alpha+\beta)}\)
Example 9.
A particle travelling with uniform acceleration along a straight line has average velocities v1, v2 and v3 in successive time intervals t1, t2 and t3, respectively. Prove that, \(\frac{v_2-v_1}{v_3-v_2}\) = \(\frac{t_1+t_2}{t_2+t_3}\)
Solution:
Let the initial velocity of the particle be u and its acceleration be a. Also, x, y and z are the velocities after the respective time intervals t1, t2 and t3.
Example 10.
A bullet, moving with a velocity of 200 m ᐧ s-1 can just go through a 4 cm thick plank. What should be the velocity of a bullet for just going through a 10 cm thick Identical plank? [HS 2000]
Solution:
The retardation (a) should be the same inside both the planks. The final velocity of the bullet in both cases is zero.
Let the initial velocities of the bullet in the two cases be u1 and u2, respectively.