Question

A particle is moving with an initial velocity of 5m/s,after 10 second it's velocity decrease to 1m/s.find the value of retardation​

Answer 1

Given :

Initial velocity = 5 m/s

Final velocity = 1 m/s

Time interval = 10 s

To Find :

Retardation of the particle.

Solution :

❖ Acceleration is defined as the rate of change of velocity.

• It is a vector quantity having both magnitude as well as direction.
• It can be positive, negative or zero.
• SI unit : m/s²
• Dimension formula : [L¹ T‾² ]

First equation of kinematics is given by

$\star\:\underline{\boxed{\bf{\red{v=u+at}}}}$

• v denotes final velocity

• u denotes initial velocity

• a denotes acceleration

• t denotes time

By substituting the given values;

$\sf:\implies\:1=5+a(10)$

$\sf:\implies\:10a=1-5$

$\sf:\implies\:10a=-4$

$\sf:\implies\:a=\dfrac{-4}{10}$

$:\implies\:\underline{\boxed{\bf{\purple{a=-0.4\:ms^{-2}}}}}$

[Note : Negative sign shows retardation.]

Answer 2

Question :-

→ A particle is moving with an initial velocity of 5m/s, after 10 second it's velocity decrease to 1m/s. find the value of retardation​

Given :-

→ Initial Velocity of the particle ( u ) = 5 m/s

→ Time interval ( t ) = 10 seconds

→ Final Velocity ( v ) = 1 m/s

To find :-

→ Value of retardation

Solution :-

→ As we know that ,

$\huge\sf { \boxed{v = u + at}$

$\sf Where , \\\\v = final\;velocity \\u = initial\;velocity\\a = acceleration\\t = time \; period$

→ Let's solve by substituting the values !

$\sf \longrightarrow 1 = 5 + a(10)\\\\\longrightarrow 1 = 5 + 10a\\\\\longrightarrow 10a = 1 - 5 \\\\\longrightarrow 10a = -4 \\\\\longrightarrow a = \dfrac{-4}{10} \\\\\longrightarrow a = -0.4 ms^{-2}$

$\huge\sf\purple{Retardation\;=-0.4ms^{-2}}$