a car is running at 72km/h is slowed down to 18km/h by the application of breaks over a distance of 20m. The deceleration of car i

2 thoughts on “a car is running at 72km/h is slowed down to 18km/h by the application of breaks over a distance of 20m. The deceleration of car i”

1. [tex]\begin{gathered}{\Huge{\textsf{\textbf{\underline{\underline{\purple{Answer:}}}}}}}\end{gathered}[/tex]

   Given :

   • A car is running at 72km/h is slowed down to 18km/h by the application of breaks over a distance of 20m.

   To Find :

   Deceleration

   Solution :

   • We know that

   • 1 km/h = 5/18 m/s

   • For Initial velocity

   • 72 × 5/18 = 4 x 5 = 20 m/s

   • For Final velocity

   • 18 × 5/18 = 1× 5 = 5 m/s

   So now, Deceleration =

   • 5²-20² = 2a(20)

   • 25 – 400 = 40a

   • -375 = 40a

   . -375/40 a

   -9.375 = a

   Acceleration = -9.375 m/s²

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2. Answer:

   u=72km/h= 72 ×5/18 =20m/s

   v=18km/s= 18×5/18 = 5m/s

   s=20m

   then,

   v^2=u^2 + 2as

   5^2=20^2 + 2a*20

   a= -375/40

   a=-75/8 = -9.375 m/s^2

   deceleration of car is 9.375 m/s*2

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