Question

a car is travelling with a speed of 36 kilometre per hour but driver applies the break and retards the car uniformly the car is stopped in 5 second find the retardation and the distance travelled before it is stops.​

Answer 1

Answer:

Retardation is 2 m/s² & Distance travelled is 25 m.

Explanation:

u = 36 kmph

⇒ u = 36 × $\frac{5}{18}$ m/s = 10 m/s

t = 5 s

As v = u + at

v = 0 (As the car stops)

⇒ 0 = u + at

⇒ u = (-a)t

⇒ 10 = (-a) × 5

∴ a = - 2 m/s² (negative sign indicates opposite direction)

As s = ut + $\frac{1}{2}$at²

⇒ s = 10(5) + $\frac{1}{2}$(-2)(5²)

⇒ s = 50 - 25

∴ s = 25 m

Hoping that I have not made any mistakes, You're welcome.

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GeniusH