car is travelling with speed of 36km/h. the driver apply brake and retars uniformly the is stopped in 5sec .now find the retardation and distance travelled before it stopped after applying brake.
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Given -
u = 36 km /h = 36× 1000m/ 1×60×60s
=> 10m/s
=> 5 sec
v = 0
Using , v = u + at
0 = 10 +a × 5
=> 5a = -10
=> a = -10/5
=> a = - 2m/s^2
Hence retardation = -2m/s^2
by 2nd equation of motion we obtained
( s = ut + 1/2 at^2 { since 2nd equation of motion }
s = 10 × 5 + 1/2 × ( -2) × 25
s = 50 + 12.5 × -2
s = 50 + (-25)
s = 25 m
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Given -
u = 36 km /h = 36× 1000m/ 1×60×60s
=> 10m/s
=> 5 sec
v = 0
Using , v = u + at
0 = 10 +a × 5
=> 5a = -10
=> a = -10/5
=> a = - 2m/s^2
Hence retardation = -2m/s^2
by 2nd equation of motion we obtained
( s = ut + 1/2 at^2 { since 2nd equation of motion }
s = 10 × 5 + 1/2 × ( -2) × 25
s = 50 + 12.5 × -2
s = 50 + (-25)
s = 25 m
please mark it as brainliest and don't forget to press heart icon
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