please guys answer this question
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Steph0303:
Is the answer 25 m ?
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Given,
u= 54km/h = 15m/sec
a = -0.3m/sec²
Now, After applying break it's final speed will be zero,
Using first euqation of motion.
v = u + at
0 = 15 - 0.3×t
t = 50sec
Hence, car will stop moving after 50sec. then after 2min it will remain at same place as after 50 sec
Now,
Using third equation of motion,
v² = u² +2as
0 = 15² + 2× 0.3 × s
225 = 0.6×s
s = 375m
Hence car will travel 375 towards check point.
Therefore it's distance from check point,
= 400 - 375
= 25 m
u= 54km/h = 15m/sec
a = -0.3m/sec²
Now, After applying break it's final speed will be zero,
Using first euqation of motion.
v = u + at
0 = 15 - 0.3×t
t = 50sec
Hence, car will stop moving after 50sec. then after 2min it will remain at same place as after 50 sec
Now,
Using third equation of motion,
v² = u² +2as
0 = 15² + 2× 0.3 × s
225 = 0.6×s
s = 375m
Hence car will travel 375 towards check point.
Therefore it's distance from check point,
= 400 - 375
= 25 m
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