Question

a car is moving with 72 kilometre per hour and after applying break it is top in 2 seconds the find the retardation?​

Answer 1

$\huge\sf\pink{Answer}$

☞ Acceleration = 10 m/s

━━━━━━━━━━━━━

$\huge\sf\blue{Given}$

✭ Initial Velocity (u) = 72 km/h

✭ Final Velocity (v) = 0 km/h

✭ Time (t) = 2 sec

━━━━━━━━━━━━━

$\huge\sf\gray{To \:Find}$

◈ The retardation (Negative acceleration)

━━━━━━━━━━━━━

$\huge\sf\purple{Steps}$

So here we shall use the first Equation of motion, that is,

$\underline{\boxed{\green{\sf v = u+at}}}$

Here the speed is given in km/h so on converting to m we have to multiply by $\sf\dfrac{5}{18}$

➳ $\sf\dfrac{72 \times 5}{18}$

➳ $\sf \dfrac{360}{18}$

➳ $\sf \red{Initial \ Velocity = 20 \ m/s}$

Substituting the given values,

➝ $\sf v = u+at$

➝ $\sf 0 = 20 + a \times 2$

➝ $\sf -20 = 2a$

➝ $\sf \dfrac{-20}{2} = a$

➝ $\sf\orange{a = -10 \ m/s}$

☯ $\underline{\sf More \ Equations}$

»» s = ut+½ at²

»» v² - u² = 2as

━━━━━━━━━━━━━━━━━━

Answer 2

Given :-

• initial velocity of Car, u = 72 km/h
• final velocity of Car, v = 0

( since it stopped after applying breaks )

• Time taken by Car to stop, t = 2 s

To find :-

• Retardation of the Car, a = ?

[ Retardation is simply called negative acceleration ]

Formula required :-

• First equation of motion

v = u + a t

[ where v is final velocity, u is initial velocity, a is acceleration and t is time taken ]

Calculation :-

Converting initial velocity given in km/h into m/s

➥ u = 72 km/h

➥ u = 72 × ( 5 / 18 ) m/s

➥ u = 20 m/s

Now,

Let, acceleration of Car be a

then,

Using first equation of motion

➥ v = u + a t

( putting known values )

➥ 0 = 20 + a ( 2 )

➥ - 20 = 2 a

➥ a = -10 m/s²

so we get,

Acceleration of Car is = -10 m/s²

Therefore,

Retardation of Car is 10 m/s² .