Question

a 1120 kg car is travelling at a rate of 80km/hr when the brakes are applied if the car stops in 70 km find the stopping force in Newton kg wt ​

Answer 1

Explanation:

mass=1120kg

initial velocity=80km/hr

distance=70km

acceleration=v-u=80

Force=ma

force=80×1120 towards the car=89600N

Answer 2

Given :-

Mass of the car = 1120 kg

Initial velocity = 80 km/hr

Distance traveled = 70 km

To Find :-

The stopping force in :- Newton Kg wt

Solution :-

We know that,

• u = Initial velocity
• m = Mass
• v = Final velocity
• d = Distance

Given that,

Mass of the car (m) = 1120 kg

Initial velocity (u) = 80 km/hr = 22.22 m/s

Distance traveled (d) = 70 km  = 0.07 m

According to the question,

First we have to the acceleration of the car.

By the third equation of motion,

$\underline{\boxed{\sf v^{2}-u^{2}=2as}}$

Substituting their values, we get

$\sf 0 - 493.7 = 2 \times a \times 70000$

$\sf a=\dfrac{-493.7}{140000}$

$\sf a = 3.5 \times 10^{-3} \ m/s^{2}$

Hence acceleration of the car is 3.5 × 10⁻³ m/s²

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

Substituting their values, we get

$\sf Force=1120 \times 3.5 \times 10^{-3}$

$\sf Force=3.92 \ N$

Therefore, the stopping force is 3.92 N