Question

a car travelling with a speed of 72km/hr. the driver applied the breaks and retarded the car uniformly. the car is stopped in 5sec. find acceleration of the car, 2-distance before it stops after applying the break​

Answer 1

✨Heya!!!!✨

here is ur answer mate....❤❤❤

u=72 km/h

v=0 km/h

t=5 sec

To find acceleration:-

v=u+at

0=72+a×5

0=72+5a

5a+72=0

5a=-72

a=-72/5

a=-14.4

To find distance:-

s=ut+1/2at^2

s=72×5 +1/2×-14.4×5×5

s=360+(-180)

s=180 km

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