J's answer to LockB's Secondary 4 A Maths Singapore question.
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Braking distance = k(speed)², where k is a constant.
When braking distance = 20m, speed = v m/s,
20 = kv²
k = 20/v²
So when the speed is v/2 m/s,
Braking distance = k(speed)²
= (20/v²)(v/2)² m
= (20/v²)(v²/4) m
= 5m
Percentage decrease = decrease/original × 100 %
= (20 - 5)/20 × 100 %
= 15/20 × 100%
= 75 %
When braking distance = 20m, speed = v m/s,
20 = kv²
k = 20/v²
So when the speed is v/2 m/s,
Braking distance = k(speed)²
= (20/v²)(v/2)² m
= (20/v²)(v²/4) m
= 5m
Percentage decrease = decrease/original × 100 %
= (20 - 5)/20 × 100 %
= 15/20 × 100%
= 75 %
Date Posted:
3 years ago
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