A car is moving at 108 km/h and takes 4 sec to stop after the brake ap

1.	A car is moving at 108 km/h and takes 4 sec to stop after the brake applied. Calculate the force acting on the car after the brake applied. Total mass of the car with passenger is 1000 kg. fast solve this question and reply me.
Answer» Given : - •Inital velocity of CAR = 108km/hr (u = 108km/hr) •Time TAKEN to stop the car after applying brakes = 4sec(t = 4sec) •Final velocity = 0 (v = 0) •Mass of the car with passenger = 1000kg(m = 1000kg) To Find : - •The Force of the car. Solution : - Given, u = 108km/hr Convert km/hr into m/sec $\sf \: 108 \frac{km}{hr} = 108 \times \frac{5}{18} \\ \sf = 6 \times 5 \\ \sf= 30 \frac{m}{sec}$ v = 0 m = 1000kg we know that, $\: \sf \: force = mass \times acceleration \\ \\ \: \sf \:force = m \times \frac{v - u}{t}$ Now substitute the value $\sf \implies \: f = 1000 \times \frac{0 - 30}{4} \\ \\ \: \sf \implies \: f = 1000 \times \frac{ - 30}{4} \\ \\ \: \sf \implies \: f = 250 \times - 30 \\ \\ \: \sf \implies \: f = - 7500 \: newton$ Hence,opposite force is demoted by NEGATIVE SIGN.

A car is moving at 108 km/h and takes 4 sec to stop after the brake applied. Calculate the force acting on the car after the brake applied. Total mass of the car with passenger is 1000 kg. fast solve this question and reply me.

Answer»

Given : -

•Inital velocity of CAR = 108km/hr (u = 108km/hr)

•Time TAKEN to stop the car after applying brakes = 4sec(t = 4sec)

•Final velocity = 0 (v = 0)

•Mass of the car with passenger = 1000kg(m = 1000kg)

To Find : -

•The Force of the car.

Solution : -

Given,

u = 108km/hr

Convert km/hr into m/sec

$\sf \: 108 \frac{km}{hr} = 108 \times \frac{5}{18} \\ \sf = 6 \times 5 \\ \sf= 30 \frac{m}{sec}$

v = 0
m = 1000kg

we know that,

$\: \sf \: force = mass \times acceleration \\ \\ \: \sf \:force = m \times \frac{v - u}{t}$

Now substitute the value

$\sf \implies \: f = 1000 \times \frac{0 - 30}{4} \\ \\ \: \sf \implies \: f = 1000 \times \frac{ - 30}{4} \\ \\ \: \sf \implies \: f = 250 \times - 30 \\ \\ \: \sf \implies \: f = - 7500 \: newton$

Hence,opposite force is demoted by NEGATIVE SIGN.