Question

A car moving with a speed of 72 km/h is brought to rest in 10 seconds
by applying brakes. Find the magnitude of average retardation d

Answer 1

Answer:

• Magnitude of retardation of Car = 2 m/s²

Explanation:

Given

• Initial velocity of Car, u = 72 km/h
• Final velocity of Car, v = 0
• Time taken for changing velocity, t = 10 s

To find

• Magnitude of average retardation =?

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 ]

Solution

Let us first convert initial velocity of Car from km/h to m/s unit

→ u = 72 km/h

→ u = 72 × ( 1000 m) / ( 3600 s )

→ u = 72 × ( 1000 / 3600 )   m/s

→ 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

→ 0 = 20 + a ( 10 )

→ 0 = 20 + 10 a

→ 10 a = - 20

→ a = -2 m/s²

Therefore,

• Magnitude of retardation of the Car would be 2 m/s².

Answer 2

AnswEr :

• Initial velocity (u) = 72 km/h = 20 m/s
• Final velocity (v) = 0 m/s
• Time (t) = 10s

We've to find retardation, let's study about retardation :

Acceleration which have opposite direction from that of force is called as retardation. Magnitude of retardation is taken as negative in sign.

Applying Newton's First equation of motion :

$\\ \longrightarrow \tt{v\ =\ u\ +\ at} \\ \\ \\ \longrightarrow \tt{0\ =\ 20\ +\ 10a} \\ \\ \\ \longrightarrow \tt{10a\ =\ -20} \\ \\ \\ \longrightarrow \tt{a\ =\ \dfrac{-20}{10}} \\ \\ \\ \large \longrightarrow \underline{\boxed{\frak{a\ =\ -2\ ms^{-2}}}}$