Question

a car start moving with a initial velocity 20m/s and uniform accerlation 2m/s sq find the displacement of the particle when the final velocity become 30m/ s​

Answer 1

Given :

▪ Initial velocity of car = 20m/s

▪ Final velocity of car = 30m/s

▪ Acceleration of car = 2m/s$^2$

To Find :

▪ Displacement of the car during the acceleratory period.

Concept :

✒ Since, acceleration of car has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this type of questions.

✒ Third equation of kinematics is given by

$\bigstar\:\underline{\boxed{\bf{\green{v^2-u^2=2as}}}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes displacement

Calculation :

$\longrightarrow\sf\:v^2-u^2=2as\\ \\ \longrightarrow\sf\:(30)^2-(20)^2=2(2)s\\ \\ \longrightarrow\sf\:900-400=4s\\ \\ \longrightarrow\sf\:500=4s\\ \\ \longrightarrow\underline{\boxed{\bf{\red{s=125m}}}}$

____________________________

▶ Additional information :

• Displacement and acceleration both are vector quantities.
• Vector quantity has both magnitude as well as direction.
• Displacement and acceleration of the moving body can be positive, negative and zero.

Answer 2

$\rule{200}4$

$\huge\tt{GIVEN:}$

• Initial velocity = 20m/s
• Uniform accerlation = 2m/s sq
• Final velocity = 30m/s

$\rule{200}2$

$\huge\tt{TO~FIND:}$

• The displacement of the car.

$\rule{200}2$

$\huge\tt{CONCEPT~USED:}$

• We know the third equation of kinematics, v² - u² = 2as
• Where , v = final speed , u = initial speed , a = acceleration , s = Distance

$\rule{200}2$

$\huge\tt{SOLUTION:}$

So, using this concept here we got,

⇝v² - u² = 2as

⇝(30)² - (20)² = 2(2)s

⇝ 900 - 400 = 4s

⇝500 = 4s

⇝500/4 = s

⇝ 125 = s

${\boxed{\fbox{\fbox{\sf{\red{DISTANCE~=~125~Meter}}}}}}$

$\rule{200}4$