Question

a car is running at a speed of 72kmph on a horizontal road. on turning the engine off it stops in 20seconds . Calculate the retardation produced .​

Answer 1

Answer:

Retardation is 1 m / sec²

Explanation:

Given :

Initial velocity ( u ) = 72 km / hr

In term of m / sec .

u = 72 × 5 / 18 = 20 m / sec

When car stopped v become zero . i.e. v = 0

Time ( t ) = 20 sec

From first equation we have :

v = u + at

Putting values here :

0 = 20 + 20 a

20 a = - 20

a = -1 m / sec².

Here negative sign denotes negative acceleration also called retardation.

Retardation is 1 m / sec²

Hence we get answer.

Answer 2

AnswEr :

⠀

$\underline{\bold{\color{Lemon}{Retardation~Produced~by~Car~is~1m/sec}}}$

⠀

Explanation :

⋆ Initial Speed (u) = 72 km/hr

Changing it into m/sec.

⋆ u = 72 × $\large\bold{\frac{5}{18}}$

⠀⠀= 4 × 5

⠀⠀= 20 m/sec

⋆ Final Speed (v) = As Car Stops at End.

⋆ v = 0 m/sec

⋆ Time (t) = 20 sec

⠀

━━━━━━━━━━━━━━━━━━━━━━━━

⠀

▣ Using First Equation of Motion :

➟ v = u + at

➟ 0 = 20 + (a × 20)

➟ 0 = 20 + 20a

➟ - 20 = 20a

➟ a = $\large\bold{\frac{-20}{20}}$

➟ a = - 1 m/sec

⠀

$\large\therefore$ Retardation of the Car is 1 m/sec.