Question

a car of mass 800kg is travelling with velocity of 20m/s .when breaks are applied it stops after traveling a distance of 8m find acceleration and force​

Answer 1

Given :

• Mass of the car = 800 kg.

• Intial Velocity of the car = 20 m/s.

• Distance covered by the car before coming to rest = 8 m.

To find :

• Acceleration of the car.

• Force exerted by the car.

Solution :

Acceleration of the car :

We know the Third Equation of Motion i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² + 2aS

Where :

• v = Final Velocity
• u = initial velocity
• a = Acceleration
• S = Distance

By using the third Equation of Motion and substituting the values in it, we get :

==> v² = u² + 2aS

==> 0² = 20² + 2 × a × 8

==> 0 = 400 + 16a

==> -400 = 16a

==> -400 = 16a

==> -(400/16) = a

==> -25 = a

∴ a = - 25 m/s².

Hence the Acceleration of the car is -25 m/s or the Retardation of the car is 25 m/s²

Force exerted by the car :

We know that ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀F = ma

Where :

• F = Force
• m = mass
• a = Acceleration

By using the formula for force and substituting the values in it, we get :

==> F = ma

==> F = 800 × (-25)

==> F = (-20,000)

∴ F = -20,000 N

Hence the force exerted is -20,000 N.

Answer 2

Given :-

Mass of the car = 800 kg.

Intial Velocity of the car, u = 20 m/s.

Distance covered by the car, s = 8 m.

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

To find :-

(i) Acceleration of the car.

(ii) Force exerted by the car.

Solution :-

(i) as we are provided with initial Velocity (u), final velocity (v) and distance we can use 3rd equation of motion.

so, using 3rd equation of motion .i.e.,

➠ v² = u² + 2as

here, v denotes final velocity, u denotes initial velocity, a denotes Acceleration and s Denotes Distance

by, substituting all the given values,

➠ 0² = 20² + (2)(a)(8)

➠ 0 = 400 + 16a

➠ -400 = 16a

➠ -400 = 16a

➠ -400/16 = a

➠ a = -25m/s²

thus, the Acceleration of the car is -25 m/s

|| Note :- If the speed is decreasing with time then accerlation is negative. the negative accerlation is called retardation ||

(ii) using Newton's 2nd law of motion .i.e.,

the rate of change of Momentum of linear Momentum of a body with time is proportional to the net external force acting on it

In order words,

➠ F = ma

here, F denotes force, m Denotes Mass and a denotes accerlation

by, substituting all the given values,

➠ F = (800)(-25)

➠ F = -20,000 N

thus, the force exerted is -20,000 N$.$

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