Question

tell this question answer​

Answer 1

• Time taken to stop the car by the application of brake , t = 4 sec

• The initial speed of the car , u = 36 km/hr

• Mass of the car is 400 kg.

• Mass of the driver is 65 kg.

Net mass, m is the sum of the masses of the car and the driver.

Net mass, m = 400 + 65

m = 465 kg

The SI unit of velocity is m/s

1 km/hr = 5/18 m/s

$36 \: \frac{km}{hr} = 36 \times \frac{5}{18} \: \frac{m}{s} \\ \\ = \frac{36}{18} \times 5 \: \frac{m}{s} \\ \\ = 2 \times 5 \: \frac{m}{s} \\ \\ u = 10 \: \frac{m}{s}$

Initial velocity, u = 10 m/s

Since the car is stopped by the application of brake, the final velocity of the car is zero.

Final velocity, v = 0

According to the first equation of motion,

v = u + at

Here a is the retarding acceleration.

$v = u+ at \\ \\ 0 = 10 + (a \times 4) \\ \\ - 10 = 4 \times \: a \\ \\ a = \frac{ - 10}{4} \\ \\ a = - \frac{5}{2} \\ \\ a = - 2.5 \: \dfrac{m}{ {s}^{2} }$

The retarding acceleration , a = - 2.5 m/s²

Retarding force , f = ma

$f = 465 \times ( - 2.5) \\ \\ f = - 1162.5 \: newton$

The magnitude of retarding force, f = 1162.5 N