Question

. A car is travelling at a speed of 72km/h, the brakes are applied so as to produce a uniform acceleration of 15m/s2. Find the distance it travels after applying brakes so that it completely
stops?​

Answer 1

HEY MATE!!

Speed in km/hr=72

Speed in km/hr=72Speed in m/s=72×5/18

Speed in km/hr=72Speed in m/s=72×5/18 = 20m/s

Speed in km/hr=72Speed in m/s=72×5/18 = 20m/sv²-u²=2as

Speed in km/hr=72Speed in m/s=72×5/18 = 20m/sv²-u²=2as0²-(20)²=2×-1×s

Speed in km/hr=72Speed in m/s=72×5/18 = 20m/sv²-u²=2as0²-(20)²=2×-1×s-400=-2s

Speed in km/hr=72Speed in m/s=72×5/18 = 20m/sv²-u²=2as0²-(20)²=2×-1×s-400=-2ss=200m

Speed in km/hr=72Speed in m/s=72×5/18 = 20m/sv²-u²=2as0²-(20)²=2×-1×s-400=-2ss=200m∴The car will travel a distance of 200m before coming to rest.

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Answer 2

ANSWER:

IT TRAVELS 16.66 m AFTER APPLYING BRAKES TILL IT STOPS

GIVEN:

u = 72 kmph = 72 × 5/18 = 20m/s

v = 0

a = - 15 m/s²

TO FIND:

distance travelled by the car after applying brakes.

FORMULA:

v² - u² = 2as

EXPLANATION:

0² - 20² = 2 × -15 × S

-400 = -30 × S

S = 40/3

S = 16.66 m

IT TRAVELS 16.66 m AFTER APPLYING BRAKES TILL IT STOPS.

VERIFICATION:

STILL WE HAVE TWO MORE FORMULAS, THEY ARE

v = u + at

s = ut + 1/2 at²

LET US USE BOTH IF THESE TO VERIFY THE ANSWER WE GOT

0 = 20 - (-15) t

-20 = - 15t

t = 4/3 seconds

s = 20(4/3) + 1/2 × -15 + 16/9

s = 80/3 - (15 × 8/9)

s = 80/3 - (5×8/3)

s = 80/3 - 40/3

s = (80 - 40) / 3

s = 40/3 = 16.66 m

HENCE OUR ANSWER IS VERIFIED.