Question

A car is moving with a velocity of 90km/hr and it applies break., after break car moves around 6m.Find the retardation of car​

Answer 1

Given:-

→ Initial velocity of the car = 90km/h

→ Distance travelled = 6m

To find:-

→ Retardation of the car.

Solution:-

Firstly, let's convert the initial velocity of the car from km/h to m/s.

=> 1km/h = 5/18 m/s

=> 90km/h = 5/18×90

=> 25m/s

Also, in this case :-

• Final velocity of the car will be zero as it

finally comes to rest after application of

brakes.

Now, by the 3rd equation of motion :-

v² - u² = 2as

Where :-

• v is final velocity of the car.

• u is initial velocity of the car.

• a is acceleration of the car.

• s is the distance travelled.

Substituting values, we get :-

=> 0 - (25)² = 2(a)(6)

=> -625 = 12a

=> a = -625/12

=> a = -52.083 m/s²

[Here, -ve sign shows retardation]

Thus, retardation of the car is 52.083m/s² .

Answer 2

We have,

Intial velocity = 90 km/h = 25 m/s

Final velocity = 0 m/s {Comes to rest}

Distance travelled = 6 m

We know,

2as = v² - u² {Position - velocity; 3rd equation of motion}

⇒ 2as = - u²

⇒ a = u²/2s

⇒ a = - (25 m/s)²/(2 × 6 m/s)

⇒ a = - (625 m²/s²)(12 m/s)

⇒ a = - 52.0833 m/s² {Answer}