if the weight of the truck is inversely proportional to the square of its speed. then by percentage does its speed decrease if their is an increase in its weight by 44% ? About the author Melanie
Given info : if the weight of the truck is inversely proportional to the square of its speed. There is an increase in its weight by 44 %. To find : The percentage decrease in the speed of truck is … solution : let weight of truck is w and speed of it is v. a/c to question, w ∝ 1/v² ⇒w = k/v² …(1), where k is proportionality constant. now w is increased by 44 % so, new weight = w + 44 % of w = 1.44 w so, 1.44w = k/v’² ⇒v’ = √(k/1.44w) ….(2) now new speed of truck, v’ = √(k/1.44w) [ from eq (2) ] now percentage change in v = (v’ – v)/v × 100 = {√(k/1.44w) – √(k/w)}/√(k/w) × 100 = (1/1.2 – 1)/1 × 100 = -0.2/1.2 × 100 = -100/6 = -16.67 % Therefore the percentage decrease in speed of truck is 16.67 % Reply
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Given info : if the weight of the truck is inversely proportional to the square of its speed. There is an increase in its weight by 44 %.
To find : The percentage decrease in the speed of truck is …
solution : let weight of truck is w and speed of it is v.
a/c to question,
w ∝ 1/v²
⇒w = k/v² …(1), where k is proportionality constant.
now w is increased by 44 %
so, new weight = w + 44 % of w = 1.44 w
so, 1.44w = k/v’²
⇒v’ = √(k/1.44w) ….(2)
now new speed of truck, v’ = √(k/1.44w) [ from eq (2) ]
now percentage change in v = (v’ – v)/v × 100
= {√(k/1.44w) – √(k/w)}/√(k/w) × 100
= (1/1.2 – 1)/1 × 100
= -0.2/1.2 × 100
= -100/6
= -16.67 %
Therefore the percentage decrease in speed of truck is 16.67 %