A motorist travelling at 10m/s was able to bring his car to rest in a distance of 10 meters. If he had been travelling at 40 m/s, in what distance could he bring his car to rest using the same braking force? (answer in meters)
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Answered by
25
initial velocity = 10 m/s
final velocity = 0 m/s
distance covered = 10 m
acceleration of the bike,
2as = v^2-u^2
2(a)(10) = (0)^2-(10)^2
20 a = -100
a = -100/20
a = -5 m/s^2
therefore in the next condition,
initial velocity = 40 m/s
final velocity = 0 m/s
acceleration = -5 m/s^2 ( as mentioned in the question above )
therefore distance traveled by the bike,
2as = v^2-u^2
2(-5)(s) = (0)^2-(40)^2
-10 s = -1600
s = -1600/-10
s = 160 m.
i hope that this will help u.
final velocity = 0 m/s
distance covered = 10 m
acceleration of the bike,
2as = v^2-u^2
2(a)(10) = (0)^2-(10)^2
20 a = -100
a = -100/20
a = -5 m/s^2
therefore in the next condition,
initial velocity = 40 m/s
final velocity = 0 m/s
acceleration = -5 m/s^2 ( as mentioned in the question above )
therefore distance traveled by the bike,
2as = v^2-u^2
2(-5)(s) = (0)^2-(40)^2
-10 s = -1600
s = -1600/-10
s = 160 m.
i hope that this will help u.
Answered by
1
Answer:
160 m is the required answer
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