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secondary 4 | A Maths
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need help with this qn, pls explain too
Date Posted: 3 years ago
Views: 395
If I call the braking distance D,
D = kv²
20 = kv² for some value of v
When the speed is reduced by 50%, its new speed is v/2
New braking distance
= k (v/2)²
= k (v²/4)
= kv²/4
= 20/4
= 5 m
Percentage decrease in braking distance
= (20 - 5) / 20 x 100%
= 75%
D = kv²
20 = kv² for some value of v
When the speed is reduced by 50%, its new speed is v/2
New braking distance
= k (v/2)²
= k (v²/4)
= kv²/4
= 20/4
= 5 m
Percentage decrease in braking distance
= (20 - 5) / 20 x 100%
= 75%
See 1 Answer
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 %
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