Question

The brakes applied to car produce an acceleration of 8 m/s2 in the opposite direction to the motion. For how long are the breaks to be applied so that its velocity is reduced from 108km/h to 18km/h.

Answer 1

▶ Solution :

Given :

▪ Deceleration of car = 8m/s^2

▪ Initial velocity = 108kmph

▪ Final velocity = 18kmph

To Find :

▪ Interval of time

Concept :

✒ Since, Acceleration is constant throughout the whole journey we can apply equation of kinematics directly.

✒ First equation of kinematics is given by

☞ v = u + at

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time interval

Conversion :

⏭ 1kmph = 5/18mps

⏭ 108kmph = 108×5/18 = 30mps

⏭ 18kmph = 18×5/18 = 5mps

Calculation :

→ v = u - at

Negative sign shows retardation

→ 5 = 30 - 8t

→ 5 - 30 = -8t

→ -25 = -8t

→ t = 25/8

→ t = 3.125s

Answer 2

Given :-

◆ Acceleration (Retardation)[a]= -8m/s^2

( because break applied )

◆ Initial velocity (u) =108km/h

◆ Final velocity (v)= 18km/h

To find :-

The time taken before the car come in rest !

$\underline{\bigstar{\sf{ By\ first\ equation\ of \ motion}}}$

$\bigstar{\boxed{\textsf{\textbf{\dag \ \ v=u+at}} }}$

• First convert Initial and final velocity of car in m/s

$\bullet\sf 1 \ km/h =\dfrac{5}{18}m/s\\ \\ \implies\sf 108km/h = \cancel{108}\times \dfrac{5}{\cancel{18}}\\ \\ \implies\sf u= 6\times 5 = 30m/s\\ \\ \implies\sf 18km/h = \cancel{18}\times \dfrac{5}{\cancel{18}}\\ \\ \implies\sf v= 5m/s$

• Put the given values In the formula

$:\implies\sf v=u+at\\ \\ :\implies\sf 5= 30+ (-8)\times t\\ \\ :\implies\sf 5-30 = -8t\\ \\ :\implies\sf \dfrac{-25}{-8}=t\\ \\ :\implies\sf 3.125=t\\ \\ :\implies\sf time (t)= 3.125 \ seconds$

$\underline{\boxed{\textsf{\textbf{\dag \ \ Time \ taken \ by\ car \ to \ come \ in \ rest = 3.125sec}} }}$