Question

a motor car is moving with a velocity of 108 kilometre per hour and it takes 4 second to stop after the break are applied. calculate the force exerted by the breaks in Motor car is if its mass along with the passenger is 1000 kg ​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

u = 108 km/h

u = 108 × 5/18 m/s.

• u = 30 m/s.
• t = 4 seconds.
• v = 0 m/s
• m = 1000 Kg.

$\huge{\bold{\underline{Explanation:-}}}$

As we need to find Force applied for that we need to find acceleration of the motor car.

From first kinematics equation.

$\large{\bold{v = u + at}}$

$\large{a = \frac{v - u}{t}}$

Substituting the values.

$\large{a = \frac{0 - 30}{4}}$

$\large{a = - 7.5 m/s^2}$

$\huge{\boxed{\boxed{a = - 7.5 m/s^2 }}}$

Now,applying Newton's second law.

$\large{\bold{F = ma}}$

Substituting the values.

$\large{ F = 1000 \times - 7.5}$

$\large{F = - 7500N}$

$\huge{\boxed{\boxed{F = - 7500N}}}$

Taking only magnitude gives.

☆ F = 7500N.

So,the Force applied by the brakes is 7500N .

The negative sign indicates that it is opposing the motion of the body.