What Are The Equations Of Motion (2024)

Equations Of Motion

Motion under uniform acceleration

1stEquation of motion
For an object moving with uniform velocity, v, its displacement, s after time, t is given by:
s = v ×t
Consider a body having initial velocity ‘u’. Suppose it is subjected to a uniform acceleration ‘a’ so that after time ‘t’ its final velocity becomes ‘v’. Now we know,
\( \text{ }\!\!~\!\!\text{ Acceleration}=\frac{\text{change}\,\,\text{in}\,\text{velocity}}{\text{Time}} \)
\( a=\frac{v-u}{t} \)
or v = u + at …..(i)

2nd Equation of motion
Suppose a body has an initial velocity ‘u’ and uniform acceleration ‘a’ for time ‘t’ so that its final velocity becomes ‘v’. The distance travelled by moving body in time ‘t’ is ‘s’ then the average velocity = (v + u)/2.
Distance travelled = Average velocity × time
\( \text{S}=\left( \frac{u+v}{2} \right)t\text{ } \)
\( \Rightarrow \text{S}=\left( \frac{u+u+at}{2} \right)t\text{ }\left( as\text{ }v=u+at \right)~ \)
\( \Rightarrow \text{S}=\left( \frac{2u+at}{2} \right)t \)
\( \Rightarrow \text{S}=\frac{2ut+a{{t}^{2}}}{2} \)
\( \text{S}=ut+\frac{1}{2}a{{t}^{2}}\text{ }……\text{ (ii)} \)

3rd Equation of motion
Distance travelled = Average velocity x time
\( S=\left( \frac{u+v}{2} \right)t\text{ }………\text{ (iii)} \)
\( \text{from equation }\left( \text{i} \right)\text{ }t=\frac{v-u}{a} \)
Substituting the value of t in equation (iii),
\( \text{we get }S=\left( \frac{v-u}{a} \right)\left( \frac{v+u}{2} \right) \)
\( S=\left( \frac{{{v}^{2}}-{{u}^{2}}}{2a} \right) \)
⇒2as = v2 – u2 or
v2 = u2 + 2as ….(iv)

The equations of motion under gravity can be obtained by replacing acceleration by acceleration due to gravity (g) and can be written as follows

  • When the body is coming towards the centre of earth
    (a) v = u + gt (b) h = ut + \(\frac { 1 }{ 2 }\) gt2 (c) v2 = u2 + 2gh
  • When a body is thrown upwards with some initial velocity, then a retardation produced due to attraction of the earth. In equations of motion, a is replaced by (–g) and thus equations become.
    (a) v = u – gt (b) h = ut – \(\frac { 1 }{ 2 }\) gt2 (c) v2= u2 – 2gh

Body Falling Freely Under Gravity

Assuming u = 0 for a freely falling body
What Are The Equations Of Motion (1)

Body is projected vertically up:
Taking initial position as origin and direction of motion (i.e. vertically up) as positive.
(a) At the highest point v = 0
(b) a = – g
What Are The Equations Of Motion (2)
It is clear that in case of motion under gravity
(a) Time taken to go up is equal to the time taken to fall down through the same distance.
(b) The speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.
(c) The body returns to the starting point with the same speed with which it was thrown.

Equations Of Motion Example Problems With Solutions

Example 1. A car is travelling with a uniform velocity of 80 km h-1 northward from Johor Bahru.
What is its displacement after 15 minutes?
What Are The Equations Of Motion (3)

Example 2. A train travelling in a straight line at 30 ms-1accelerates uniformly to 54 ms-1 in 3.0 seconds. Calculate the distance travelled by the train during that time.
What Are The Equations Of Motion (4)

Example 3. A school bus accelerates with an acceleration of 4.0 m s-2after picking up some students at a bus stop.
What Are The Equations Of Motion (5)
Calculate the
(a) velocity of the bus after 5 s.
(b) distance travelled by the bus after 5 s.
Initial velocity, u = 0 ms-1
Acceleration, a = 4.0 ms-2
Time, t = 5 s
What Are The Equations Of Motion (6)

Example 4.A long jumper was running at a velocity of 5 m s-1towards the long jump pit. He needed to achieve a velocity of 10 m s-1after covering a distance of 4.5 m before lifting himself off the ground from the jumping board.
What Are The Equations Of Motion (7)
(a) Calculate the required acceleration for him to do so.
(b) Calculate the time taken for him to cover the horizontal distance of 4.5 m.
What Are The Equations Of Motion (8)
Example 5.10 A body starts moving with an initial velocity 50 m/s and acceleration 20 m/s2. How much distance it will cover in 4s ? Also, calculate its average speed during this time interval.
Solution: Given: u = 50 m/s, a = 20 m/s2, t = 4s, s = ?
s = ut + \(\frac { 1 }{ 2 }\) at2 = 50 × 4 + \(\frac { 1 }{ 2 }\) × 20 × (4)2
= 200 + 160 = 360 m
Average speed during this interval,
\(\overline{V}=\frac{\text{distance travelled}}{\text{time interval}}=\frac{360}{4}=90\text{ m/s}\)

Example 6.A body is moving with a speed of 20 m/s. When certain force is applied, an acceleration of 4 m/s2 is produced. After how much time its velocity will be 80 m/s ?
Solution: Given: u = 20 m/s, a = 4 m/s2,v = 80 m/s, t = ?
Using equation, v = u + at, we get
80 = 20 + 4 × t
or 4t = 80 – 20 = 60
or t = 15 s
Therefore, after 15 seconds, the velocity of the body will be 80 m/s.

Example 7.A body starts from rest and moves with a constant acceleration. It travels a distance s1in first 10 s, and a distance s2 in next 10 s. Find the relation between s2and s1.
Solution: Given : u = 0, t1 = 10 s
∴ Distance travelled in first 10 seconds, is given by
s1 = ut + \(\frac { 1 }{ 2 }\) at2 = 0 + \(\frac { 1 }{ 2 }\) × a × (10)2
s1 = 50a …(1)
To calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s.
s = ut + \(\frac { 1 }{ 2 }\) at2 = 0 + \(\frac { 1 }{ 2 }\) × a × (20)2
s = 200a …(2)
∴ Distance travelled in 10th second interval,
s2 = s – s1 = 200a – 50a …(3)
or s2 = 150a
\(\text{Now, }\frac{{{s}_{2}}}{{{s}_{1}}}=\frac{150a}{50a}=\frac{3}{1}\)
or s2 = 3s1

Example 8.A train is moving with a velocity 400 m/s. With the application of brakes a retardation of 10 m/s2 is produced. Calculate the following
(i) After how much time it will stop ?
(ii) How much distance will it travel before it stops ?
Solution: (i) Given: u = 400 m/s, a = –10 m/s2, v = 0, t = ?
Using equation, v = u + at, we get
0 = 400 + (–10) × t
or t = 40 s
(ii) For calculating the distance travelled, we use equation,
v2 = u2 + 2as, we get
(0)2 = (400)2 + 2 × (–10) × s
or 20s = 400 × 400
or s = 8000 m = 8 km

Example 9.A body is thrown vertically upwards with an initial velocity of 19.6 m/s. If g = –9.8 m/s2. Calculate the following
(i) The maximum height attained by the body.
(ii) After how much time will it come back to the ground ?
Solution: (i) Given: u = 19.6 m/s, g = –9.8 m/s2, v = 0, h = ?
Using equation v2 = u2 + 2gh, we get
(0)2 = (19.6)2 + 2(–9.8) × h
\(h=\frac{19.6\times 19.6}{2\times 9.8}=19.6\text{ m}\)
(ii) Time taken to reach the maximum height can be calculated by the equation,
v = u + gt
0 = 19.6 + (–9.8) × t
t = 2s
In the same time, it will come back to its original position.
∴ Total time = 2 × 2 = 4s

Example 10.From the top of a tower of height 490 m, a shell is fired horizontally with a velocity 100 m/s. At what distance from the bottom of the tower, the shell will hit the ground ?
Solution: We know that the horizontal motion and the vertical motion are independent of each other. Now for vertical motion, we have u = 0,
h = 490 m, g = 9.8 m/s2, t = ?
Using equation, h = ut + \(\frac { 1 }{ 2 }\) gt2, we get
490 = 0 + \(\frac { 1 }{ 2 }\) × 9.8 × t2
or t2 = \(\frac { 490 }{ 4.9 }\) = 100
or t = 10 s
∴ It takes 10 seconds to reach the ground.
Now, horizontal distance= horizontal velocity × time
= 100 m/s × 10 s = 1000 m
∴ The shell will strike the ground at a distance of 100 m from the bottom of the tower.

What Are The Equations Of Motion (2024)


What Are The Equations Of Motion? ›

The three equations are, v = u + at. v² = u² + 2as. s = ut + ½at²

What is the 3 equation of motion? ›

The three equations are, v = u + at. v² = u² + 2as. s = ut + ½at²

What are the 4 equations of motion? ›

Any of four equations that apply to bodies moving linearly with uniform acceleration (a). The equations, which relate distance covered (s) to the time taken (t), are: v = u + at s = (u + v)t/2 s = ut + at2/2 v2 = u2 + 2as where u is the initial velocity of the body and v is its final velocity.

What are the 5 basic motion equations? ›

The equations are as follows: v=u+at,s=(u+v2)t,v2=u2+2as,s=ut+12at2,s=vt−12at2.

What is the first equation of motion? ›

The first equation of motion is v = u +at. Here, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time. The velocity-time relation gives the first equation of motion and can be used to find acceleration.

What is the equation 3 law of motion? ›

Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts. Mathematically, if a body A exerts a force →F on body B, then B simultaneously exerts a force −→F on A, or in vector equation form, →FAB=−→FBA.

What does the 3 law of motion state? ›

Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first. His third law states that for every action (force) in nature there is an equal and opposite reaction.

Is there an equation for motion? ›

Newton's second law, which states that the force F acting on a body is equal to the mass m of the body multiplied by the acceleration a of its centre of mass, F = ma, is the basic equation of motion in classical mechanics.

What are the 4 basic types of motion? ›

Now, let us talk about the distinctive types of motions that may be seen in an object. These are Linear, Rotary, Oscillating motion, and Periodic motion.

What are 5 examples of motion in physics? ›

Running, cycling, jumping, swimming, eating, drinking, playing, writing, typing, moving cars, and throwing a ball are all examples of motion. Sleeping, sitting, standing, lying, a fixed clock, a bottle on a table, and a stopped car are all examples of rest.

What is the correct equation of motion? ›

v=u+at is the first equation of motion. So option d is correct.

What is the second equation of motion? ›

The second equation of motion gives the position-time relation, i.e. s = ut + (1/2) at2.


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