Question

❤❤A driver , having a definite reaction time is capable of stopping his car over a distance of 30m on seeing a red traffic signal when the speed of the car is 72 km/hr and over a distance of 10 M when the speed is 36 km/hr. find the distance over which he can stop the car if it were running at a speed of 54 km/hr . a GM Diet his reaction time and the the deceleration of the car remains same in all the three cases.

Answer 1

Answer :-


$solution$
Let 't' second be the reaction time of the driver to be able to put breaks and let 'a' be the acceleration of the car, and full brakes are applied.

(I) distance travelled by the car comma after the driver has seen the signal and before the driver has just apply the brakes.

= velocity * time

= 72 km /hr * t

= 72 * 5/18* t m/sec

= 20 t m/sec

conversion in m/sec

72 km/hr = 20 m/sec

54 km/hr = 15 m/sec

36 km/hr = 10 m/sec


Thus the brakes stopped the car within the distance of (30-20) m

a = 0² - 20²/2(30- 20t )

= -20²/2 * (30 - 20t)

= – (20/3-2t)

similarly,

a = (0- 10²) /2(10-10t)

= -5/1-t

similarly,

a = (0-15²) /2(X - 15 t )

when X : distance travelled by car in third case

20/3-2t = 5/1-t = 15²/2(X - 15t)

t = 1/2 sec

X = 18.75 m.

____________________
❤BE BRAINLY ❤

Answer 2

Explanation:

72km/h*t

72*5/18*t m/s

=20t m/s

conversion in m/s

72km/h= 20m/s

54km/h= 15m/s

36km/h= 10m/s

thus,the brake stopped the car within the distance (30-20)

a power 2-20power 2/2(30-20t)

-20 power 2/2*(30-20t)

=-(20/3-2t)

similarly

a=(0-10power 2)/2(10-10t)

=-5/1-t

similarly

a=(0-15 power 2)/2(X-5t)

when X: distance travelled by car in third case

20/3-2t=5/1=15 power 2/2(X-15t)

t=1/2sec

X=18.75m

hope it help you