Question

A car increase its speed from10m/s to 20m/s in 10 second it's acceleration is

Answer 1

Given :

➛ Initial speed of car = 10m/s

➛ Final speed of car = 20m/s

➛ Time interval = 10s

To Find :

➾ Acceleration of car.

Concept :

➳ Acceleration is defined as the rate of change in speed.

➳ It is a vector quantity.

➳ It has both magnitude as well as direction.

➳ It can be positive, negative or zero.

➳ SI unit : m/s²

Calculation :

⇒ a = (v - u) / t

⇒ a = (20 - 10) / 10

⇒ a = 10/10

⇒ a = 1m/s²

Answer 2

$\bigstar$ $\sf{\red{\underline{\underline{\purple{CONCEPT:-}}}}}$

☞ Equation Of Motion,

• $\tt{\:v\:=\:u\:+\:at\:}$

• $\tt{\:s\:=\:ut\:+\:{\dfrac{1}{2}}at^2\:}$

• $\tt{\:v^2\:-\:u^2\:=\:2as\:}$

☞ This question is based on Basic Concept , i.e. “Rate of change in velocity/speed is known as Acceleration” .

☞ FORMULA:-

• $\tt{\:v\:=\:u\:+\:at\:}$

$\bigstar$ $\sf{\red{\underline{\underline{\purple{To\:Find:-}}}}}$

• Acceleration (a) of Car .

$\bigstar$ $\sf{\red{\underline{\underline{\purple{SOLUTION:-}}}}}$

☞ GIVEN:-

• initial Velocity (u) = 10 m/s
• final Velocity (v) = 20 m/s
• time (t) = 10 second

☞ Now we have know that,

$\tt{\:v\:=\:u\:+\:at\:}$

$\tt{\implies\:a\:=\:{\dfrac{v\:-\:u}{t}}}$

$\tt{\implies\:a\:=\:{\dfrac{20\:-\:10}{10}}}$

$\tt{\implies\:a\:=\:{\dfrac{10}{10}}}$

$\tt{\implies\:a\:=\:1\:m/s^2}$

☞ Therefore, Acceleration of Car is

“1 m/s²” .

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::1\:m/s^2\:}}}}$