Question

a car running at a speed of 72km/h is slowed down to 18km/h over a distance of of 40metres.calculate the retardation produced by the brakes and time so taken?
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Answer 1

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Answer 2

Given :

• initial velocity of Car, u = 72 km/h = 72 × ( 5/18 )  m/s = 20 m/s
• final velocity of Car, v = 18 km/h = 18 × ( 5/18 )  m/s = 5 m/s
• distance covered for velocity change, s = 40 m

To find :

• Retardation produced = ?
• time taken for changing velocity, t = ?

Formula required :

• first equation of motion

     v = u + a t

• Third equation of motion

   2 a s = v² - u²

[ v is final velocity, u is initial velocity, a is acceleration , t is time taken , s is distance covered ]

Solution :

Retardation is called as negative acceleration. so,

Let, acceleration of Car = a

then,

Using third equation of motion

→ 2 a s = v² - u²

→ 2 a ( 40 ) = ( 5 )² - ( 20 )²

→ 80 a = 25 - 400

→ a = ( - 375 ) / 80

→ a = -75/16 m/s²

• since, acceleration of Car is -75/16 m/s²
• therefore, retardation of Car is 75/16 m/s².

Using first equation of motion

→ v = u + a t

→ 5 = 20 + ( -75/16 ) t

→ 5 - 15 = ( -75/16 ) t

→ -10 / ( - 75/ 16 ) = t

→ t = 2.134 s

• therefore, Time taken for changing velocity is 2.134 seconds.