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Derivation of equations of motion from Velocity-Time (V-T)
graph
𝐶

2)𝑠 = 𝑢𝑡 + 𝑎𝑡 2

Velocity

Equations of motion are
1)𝑣 = 𝑢 + 𝑎𝑡
3)𝑣 2 = 𝑢2 + 2𝑎𝑠

𝐷

1
2

𝑎𝑡

A

𝑡

P

Here, a body is moving with uniform
acceleration 'a' in a straight line
with initial velocity 'u' at time t=0
...


dh

or, 𝑎 =

𝐷𝐵
𝐴𝐶−𝐴𝐵

𝐴 Time

Fig: Velocity-Time graph

ya

or, 𝑎 =

𝐶𝐵

𝑡

𝑂

ya

or, 𝑎 =

𝑣

𝑢

pa

U

1)𝐹𝑜𝑟 𝑣 = 𝑢 + 𝑎𝑡
-𝐹𝑟𝑜𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒 𝑔𝑟𝑎𝑝ℎ

...
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑏𝑜𝑑𝑦 = 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑙𝑖𝑛𝑒 𝐶𝐷

or, 𝑠 = 𝐴𝑟𝑒𝑎 𝑜𝑓 △ 𝐶𝐵𝐷 + 𝐴𝑟𝑒𝑎 𝑜𝑓 ▭𝑂𝐴𝐵𝐷
1

or, 𝑠 = 𝐵𝐶 × 𝐷𝐵 + 𝑂𝐴 × 𝑂𝐷
2
1

or, 𝑠 = 𝑎𝑡 × 𝑡 + 𝑢 × 𝑡
2

1


...
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑏𝑜𝑑𝑦 = 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑙𝑖𝑛𝑒 𝐶𝐷
or, 𝑠 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝑂𝐴𝐵𝐶𝐷
2

𝐶𝐵
𝑎

U

or, 2𝑠 = (𝑂𝐷 + 𝐴𝐶)

P

or, 𝑠 = (𝑂𝐷 + 𝐴𝐶 )𝐷𝐵

ya


...
Path followed
by projectile is called trajectory
...
After certain time 't' it reaches to the
point p(x,y) which covers horizontal distance 'x' and vertical distance
'y'
...
After then its velocity increase till it hit the ground in
y axis
...


Equations of trajectory:
1) Path followed by projectile is parabolic in nature
...

At 'x' direction,
1


...
𝑠𝑦 = 𝑢𝑦 𝑡 + 𝑎𝑦 𝑡 2
2

P

1

A

At 'y' direction

U

1

or, 𝑦 = 𝑢𝑠𝑖𝑛𝜃𝑡 − 𝑔𝑡 2 [∴ 𝑎𝑦 = −𝑔]
2

𝑢2 𝑐𝑜𝑠 2 𝜃

or, 𝑦 = 𝑥𝑡𝑎𝑛𝜃 − 𝑔

…………
...
∴

𝐻𝑚𝑎𝑥 =

𝑢2 𝑠𝑖𝑛2 𝜃
2𝑔

ya

ya

Now, equation 2 is the equation of parabola
...


3)Time of flight: It is the time when projectile remains in the air
...
𝑠𝑦 = 𝑢𝑦 𝑡 + 𝑎𝑦 𝑡 2
2

1

or, 0 = 𝑢𝑠𝑖𝑛𝜃𝑇 − 𝑔𝑇 2 [𝑡 = 𝑇]
2

1

or, 𝑔𝑇 2 = 𝑢𝑠𝑖𝑛𝜃𝑇
2

2𝑢𝑠𝑖𝑛𝜃


...

1

Using equation of motion, 𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2

or, 𝑅 = 𝑢𝑐𝑜𝑠𝜃𝑇 + 0

ya

𝑔

𝑔

ya

∴𝑅=

𝑢2 𝑠𝑖𝑛2𝜃

2𝑢𝑠𝑖𝑛𝜃

dh

or, 𝑅 = 𝑢𝑐𝑜𝑠𝜃 ×

pa

U

1


...
Maximum value of sin is 1
...
∴ 𝜃 = 45°

Velocity of projectile at any instant of time: At any instant of time 't'
projectile reaches to the point 𝑝′ (𝑥 ′ , 𝑦′ )
...

Now,
Using equation of motion, 𝑣 = 𝑢 + 𝑎𝑡
Here, 𝑣𝑥 = 𝑢𝑥 + 𝑎𝑥 𝑡


...


∴ 𝑣𝑦 = 𝑢𝑦 − 𝑔𝑡

A

P

Now, Resultant velocity (𝑣) = √𝑣𝑥 2 + 𝑣𝑦 2

pa

U
Then,tan 𝛽 =

ya

𝑣𝑥
−1 𝑣𝑦
tan ( )
𝑣𝑥

ya


...


Vertically fired projectile from height: When a projectile is fired from
certain height 'H' its maximum height is throwing point 'H'
...
After certain time 't' it reaches to the
point p(x,y) which covers horizontal distance 'x' and vertical distance 'y'
...
At
the point of projection its vertical component of velocity becomes zero
...

Here,
After certain time 't' it reaches to the point p(x,y) which covers
horizontal distance 'x' and vertical distance 'y'
...
𝑠𝑥 = 𝑢𝑥 𝑡 + 𝑎𝑥 𝑡 2
2

or, 𝑥 = 𝑢𝑐𝑜𝑠𝜃𝑡 + 0 [∴ 𝑎𝑥 = 0]
-∴ 𝑡 =

𝑥
𝑢𝑐𝑜𝑠𝜃

… … … … … (1)

At 'y' direction
1


...
(2) [using equation 1 and solve]

P

A

Now, equation 2 is the equation of parabola
...


U

2)Time of flight:It is the time when projectile remains in the air
...
∴ 𝑇 = √

𝑔

3) Horizontal Range: Horizontal distance covered by projectile during
its time of flight is called Range
...
𝑠𝑥 = 𝑢𝑥 𝑡 + 𝑎𝑥 𝑡 2
2

or, 𝑅 = 𝑢𝑇 + 0
∴𝑅 =𝑢×√

2𝐻
𝑔

Velocity of projectile at any instant of time: At any instant of time 't'
projectile reaches to the point 𝑝(𝑥, 𝑦)
...

Now,
Using equation of motion, 𝑣 = 𝑢 + 𝑎𝑡
Here, 𝑣𝑥 = 𝑢𝑥 + 𝑎𝑥 𝑡


...


P

A

∴ 𝑣𝑦 = 𝑔𝑡

Now, Resultant velocity (𝑣) = √𝑣𝑥 2 + 𝑣𝑦 2 = √𝑢2 + (𝑔𝑡)2

pa

U
𝑣𝑥
−1 𝑣𝑦
tan ( )
𝑣𝑥

ya


...




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