Question

a car of mass 1000kg travelling at 10m/s is brought to rest over a distance of 20m find the braking force​

Answer 1

Given :-

Mass of the car = 1000 kg

Final velocity of the car = 0 m/s

Initial velocity of the car = 10 m/s

Distance covered by the car = 20 m

To Find :-

The braking force.

Analysis :-

Here we are given with the mass, distance, final and initial velocity of the car.

Firstly find the acceleration by substituting the given values from the question using the third equation of motion.

Then find the force by substituting the values we got such that force is equal to mass multiplied by acceleration.

Solution :-

We know that,

• s = Displacement
• t = Time
• a = Acceleration
• u = Initial velocity
• m = Mass
• v = Final velocity

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2+u^2=2as}}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 10 m/s

Displacement (s) = 20 m

Substituting their values,

⇒ 0² + 10² = 2 × a × 20

⇒ 0 + 100 = 40 × a

⇒ a × 40 = 100

⇒ a = 100/40

⇒ a = 2.5 m/s

Using the formula,

$\underline{\boxed{\sf Force=Mass \times Acceleration}}$

Given that,

Mass (m) = 1000 kg

Acceleration (a) = 2.5 m/s

Substituting their values,

⇒ f = ma

⇒ f = 1000 × 2.5

⇒ f = 2500 N

Therefore, the braking force is 2500 N.

Answer 2

formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion : v² = u²-2as}}$

Given:-

• v = 0 m/s

• u = 10 m/s

• s = 20 m

Solution:-

⇒ 0² + 10² = 2 × a × 20

⇒ 0 + 100 = 40 × a

⇒ a × 40 = 100

⇒ a = 100/40

⇒ a = 2.5 m/s

formula,

$\underline{\boxed{\sf Force=Mass \times Acceleration}}$

Given

• m = 1000 kg

• a = 2.5 m/s

Solution

⇒ f = ma

⇒ f = 1000 × 2.5

⇒ f = 2500 N

Therefore, the braking force is 2500 N.