Question

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A car was moving at a rate of 18km/h .When the brakes were applied,it comes to rest in a distance of 100m.Calculate the retardation produced by the brake.


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Answer 1

Answer:

⅛ m/s²

Explanation:

18 km/h 5/18 = 5m/s

3rd equation of motion

v² = 0² - 2aS

0= (5) ² - 2×a×100

25= 200×a

a=

see in the picture

Answer 2

Answer:

0.125 m/s² is the retardation produced by the brakes.

Explanation:

Given :

• A car was moving at a rate of 18 Km/h.
• When the brakes were applied, it comes to rest in a distance of 100 m.

To find :

• Retardation produced by the brakes.

Solution :

We have,

=> v = 0 m/s

∵ Since the car comes to rest at a particular distance.

=> u = 18 Km/h

=> u = 18 × 5/18 m/s

=> u = 5 m/s

=> S = 100 m

Using Kinematics,

>> v² - u² = 2as

[3rd equation of motion]

>> 0² - 5² = 2(a)(100)

>> -25 = 200a

>> a = -25/200

>> a = -1/8

>> a = -0.125 m/s²

or,

>> Retardation produced by the brakes is 0.125 m/s².

Learn more :

• Retardation : It is nothing but negative acceleration. When a body is decreasing its velocity, that means when coming to rest, then the body is said to be retarded, or when the velocity of the body decreases then the body is said to be retarding, it's units are same as acceleration, that is m/s², but there will a be a negative sign (-), -m/s².
• First equation of motion : v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, t is the time taken.
• Second equation of motion : S = ut + 1/2 at², where S is the distance, u is the initial velocity, a is the acceleration , t is the time taken.