Question

The stopping distance d of an automobile is directly proportional to the square of its speed s. On one road, a car requires 75 feet to stop from a speed of 30 miles per hour. How many feet does the car require to stop from a speed of 48 miles per hour on the same road?

Answer 1

Answer:

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road

Step-by-step explanation:

• Direct proportion means that two quantities increase or decrease in the same ratio

• If y is directly proportional to x (y ∝ x) , then $\frac{y_{1}}{y_{2}}=\frac{x_{1}}{x_{2}}$  OR  y = k x, where k is the constant of proportionality

∵ The stopping distance d of an automobile is directly

   proportional to the square of its speed s

- That means d ∝ s²

∴  $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

∵ A car requires 75 feet to stop from a speed of 30 miles per hour

∴ d = 75 feet

∴ s = 30 miles/hour

- Change the mile to feet

∵ 1 mile = 5280 feet

∴ 30 miles/hour = 30 × 5280 = 158400 feet/hour

∵  The car require to stop from a speed of 48 miles per hour

    on the same road

- Change the mile to feet

∴ 48 miles/hour = 48 × 5280 = 253440 feet/hour

∵  $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

- Substitute the values of $d_{1}$ by 75 feet, $s_{1}$ by 158400 feet/hour

   and $s_{2}$ by 253440 feet/hour

∴ $\frac{75}{d_{2}}=\frac{(158400)^{2}}{(253440)^{2}}$

∴  $\frac{75}{d_{2}}=\frac{25}{64}$

- By using cross multiplication

∴ 25 × $d_{2}$ = 75 × 64

- Divide both sides by 25

∴ $d_{2}$  = 192 feet

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road