A car is travelling with a speed of 36 km/h . The driver applied the breaks and retards the car uniformly. The car is stopped in 5 sec. Find (i) The acceleration of car and (ii) Distance before it stops after Applying breaks?
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Answered by
10
v=0
u=36km/h=10m/s
t=5s
v=u+at
0=10+a(5)
a= -10/5= -2m/s2
s=ut+1/2at2
=10(5)+1/2(-2)(5)(5)
=50+(-25)=25m
Pls mark as brainliest :D
u=36km/h=10m/s
t=5s
v=u+at
0=10+a(5)
a= -10/5= -2m/s2
s=ut+1/2at2
=10(5)+1/2(-2)(5)(5)
=50+(-25)=25m
Pls mark as brainliest :D
Answered by
1
u=36km/h=36x5/18 m/s=10m/s
a=v-u/t
a=0-10/5
a=-2m/sec.square
a=v-u/t
a=0-10/5
a=-2m/sec.square
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