Question

A object moving with a velocity of 108km/h applied brakes and came to rest in 48 sec find the derivation and distance tracelled

Answer 1

Given :

➛ Initial velocity = 108kmph

➛ Time of journey = 48s

➛ Final velocity = zero

To Find :

➾ Deceleration.

➾ Distance travelled.

Concept :

➜ Applied brakes produces constant retardation in object.

➜ Here, Acceleration is constant throughout the motion, We can apply equation of kinematics.

◕ First equation of kinematics :

✭ v = u + at

◕ Third equation of kinematics :

✭ v² - u² = 2as

Conversion :

➳ 1kmph = 5/18mps

➳ 108kmph = 108×5/18 = 30mps

Calculation :

➠ Deceleration :

⇒ v = u + at

⇒ 0 = 30 + a(48)

⇒ a = -30/48 = -5/8

⇒ a = -0.625 m/s²

➠ Distance travelled :

⇒ v² - u² = 2as

⇒ (0)² - (30)² = 2(-0.625)s

⇒ 900 = 1.25s

⇒ s = 900/1.25

⇒ s = 720m

Answer 2

Given:-

A object moving with a velocity of 108 km/hr applying brakes and came to rest in 48.

To Find:-

Find the derivation and distance travelled.

Concept:-

ꔷApplied brakes produced constant in retardation .

ꔷ Here acceleration is also constant.Thus, we can apply kinematics equation ,

★First equation of kinematics is

V= u + at

★Third equation of kinematics is,

${\tt{v^2 - u^2 = 2as}}$

Conversion:-

${\tt\dashrightarrow{1 kmph = \frac{5}{18}mps}}$

${\tt\dashrightarrow{108 kmph = \cancel\frac{5}{18}mps = 30 mps }}$

Calculation:-

Derivation

v = u + at

➡0 = 30 + a(48)

➡a = -30/48

➡a= -5/8

➡a= - 0.625 m/s^2

Distance travelled

$\large\leadsto{\tt{v^2-u^2= 2as}}$

$\large\leadsto{\tt{(0)^2-(30)^2= 2(-0.625)s}}$

$\large\leadsto{\tt{900=1.25}}$

$\large\leadsto{\tt{s = 900/12.5}}$

$\large\leadsto{\tt{s = 720m}}$