Question

A car of mass 900 kg is travelling with a velocity of 10m/s. When brakes are applied it stops after travelling a distance of 10m. Find retarding force​

Answer 1

Given :

Mass of the car, m = 900 kg

Initial velocity of the car, u = 10 m/s

Final velocity of the car, v = 0 m/s

Distance covered by the car, s = 10 m,

To Find :

retarding Force of the car

Solution :

first we have to find accerlation as we are provided with initial Velocity, final velocity and distance covered we can use 3rd equation of motion .i.e.,

➠ v² - u² = 2as

➠ (0)² - (10)² = (2)(a)(10)

➠ -100 = 20a

➠ - 100/20 = a

➠ a = - 5 m/s²

thus, the acceleration is - 5 m/s² retardation is 5m/s²

Remember !

If the speed is decreasing with time then accerlation is negative. The negative accerlation is called retardation${.}$

Now, using Newton's 2nd law,

the rate of change of linear Momentum of the body is proportional to the net external force acting on it .i.e.,

➠ F = ma

➠ F = (900)(5)

➠ F = 4500 N

thus, the retarding force of the car is 4500N.

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Answer 2

Solution :-

As per the given data ,

• Mass of the car = 900 kg
• Initial velocity (u) = 10 m/s
• Final velocity (v) = 0 m/s ( stops )
• Distance traveled (s) = 10 m

First we need to find the acceleration of the car (a) .

As the car is moving with uniform acceleration throughout it's motion we can use the third equation of motion in order to find a

We know that ,

➜ v² = u² + 2as

On rearranging ,

➜ a = v² - u² / 2s

Now let us substitute the given values in the  above equation ,

➜ a = 0 - 100 / 2 x 10

➜ a = - 100 / 20

➜ a = - 5 m/s ²

Note : Here negative sign denotes retardation

By applying newton's second law of motion ,

➜ F = ma

➜ F = 900 x - 5

➜ F = - 4,500 N

➜ | F | = 4,500 N

The magnitude of retarding force acting on the car is 4,500 N