Question

the break applied to a car produce a uniform retardation of 1.2m/s. before stopping the car covers a distance of 12mm. how long were the break applied. ​

Answer 1

GIVEN :

acceleration, a = 1.2 m/s

Distance, s = 12 mm.

Initial velocity, u = 0.

TO FIND :

How long were the break applied by the car.

FORMULA :

Second equation of motion :-

$S \: = \; ut \: + \: \dfrac {1}{2} \: at^{2}$

SOLUTION :

Using second equation of motion,

$S \: = \: ut \: + \: \dfrac {1}{2} \: at^{2}$

→ $12 \: = \: 0 \: + \: \dfrac {1}{2} \: \times \: \dfrac {1.2}{10} \: \times \: t^{2}$

→ $\dfrac {12 \times 10}{6} \: = \: t^{2}$

→ $t^{2} \: = \: 20$

→ $\sf \boxed {t \: = \: 255 \: sec \: or \: 4.25 \: min}$

SECOND EQUATION OF MOTION :-

• Second equation of motion is $\sf S \: = \: ut \: + \: \dfrac {1}{2} \: at^{2}$
• It is used to find the distance, times and acceleration.