A car travels with a uniform velocity of 25m/s for 5s.The brakes are then applied and the car is uniformly retarded and comes to rest in further 10s.Find (i)the distance which the car travels before the brakes are applied (ii)the retardation and (iii) the distance travelled by the car after applying the brakes.
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Hi, here is ur answer.
1. distance = speed × time
s = 25 × 5
s = 125 m.
2. by Newton's first law we can write
a = (v–u)/t
a = (0–25)/10
a = –2.5 m/s^2
here –ve sign indicates that the acceleration is in opposite direction to the motion of body.
i.e. it is retardation.
3. we can use here Newton's third law of motion.
v^2 – u^2 = 2as
0^2 – 25^2 = 2 ×(–2.5) × s
–625 = –5×s
s = 125m.
please mark it as brainliest. :)
1. distance = speed × time
s = 25 × 5
s = 125 m.
2. by Newton's first law we can write
a = (v–u)/t
a = (0–25)/10
a = –2.5 m/s^2
here –ve sign indicates that the acceleration is in opposite direction to the motion of body.
i.e. it is retardation.
3. we can use here Newton's third law of motion.
v^2 – u^2 = 2as
0^2 – 25^2 = 2 ×(–2.5) × s
–625 = –5×s
s = 125m.
please mark it as brainliest. :)
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