Question

A car stoped after travelling distance 8 meters due to applying breaks at speed of 40 meters per sec.find the acceleration and retoroion of car in that period?

Answer 1

GiveN :

• Distance Travelled (s) = 8 m
• Final velocity (v) = 0 m/s
• Initial velocity (u) = 40 m/s

To FinD :

• Acceleration

SolutioN :

Use 3rd equation of Kinematics :

⇒v² - u2 = 2as

⇒0² - 40² = 2 * a * 8

⇒-1600 = 16a

⇒a = -1600/16

⇒a = -100

$\therefore$ Acceleration produced is - 100 m/s²

Answer 2

Answer :

›»› The Acceleration and retardation of car = -100 m/s²

Given :

• Distance travelled by car = 8 m
• Initial velocity of car = 40 m/s
• Final velocity of car = 0 m/s

To Find :

• Acceleration of car = ?

Required Solution :

Before we start our solution, let's first we know about Acceleration.

The rate of change of the velocity of an object is known as Acceleration.

• The SI unit of Acceleration is m/s².
• The GGS unit of Acceleration is cm/s².
• Acceleration is represented by a.

Let's solve this question...

We can find Acceleration of the wind by using the first equation of motion which says, v² = u² + 2as

Here,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• a is the Acceleration in m/s².
• s is the Distance travelled in m.

So let's find Acceleration (a) !

→ v² = u² + 2as

→ 0² = 40² + 2 × a × 8

→ 0 = 1600 + 2a × 8

→ 0 = 1600 + 16a

→ 0 - 1600 = 16a

→ -1600 = 16a

→ a = -1600/16

→ a = -100

【NOTE : -ve sign indicates the retardation of bus.】

║Hence, the Acceleration of bus is -100 m/s².║