Question

.A car moving at 80kmh is subjected to a
uniform retardation and brought to rest in
70s. Find the retardation and the distance
travelled by the car before it comes to rest.
(Ans:- 777.8 m)​

Answer 1

Given:-

• Initial velocity , u = 80km/h

• Final velocity , v = 0m/s

• Time taken ,t = 70s

To Find:-

• Retardation and distance covered by the car.

Solution:-

Firstly we convert the unit here

80km/h = 80×5/18 = 200/9 m/s

Now , by Using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 0 = 200/9 + a ×70

→ -200/9 = a×70

→ -22.22 = a ×70

→ a = -22.22/70

→ a = -0.31m/s²

Here negative sign show retardation

Therefore, the retardation of the car is 0.31m/s²

Now , calculating distance

Using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ 0² = (22.22)² + 2×(-0.31) ×s

→ 0 = 493.72 + (-0.62)×s

→ -493.72 = -0.62 ×s

→ s = -493.72/0.62

→ s = 796.33m

Therefore, the distance covered by the car is 796.33 m.

Answer 2

Answer:

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