Question

Suppose two cars depart from a four-way intersection at the same time, one heading north and the other heading west. The car heading north travels at the steady speed of 40 ft/sec and the car heading west travels at the steady speed of 66 ft/sec. (a) Find an expression for the distance between the two cars after t seconds. (Round your coefficients to one decimal place as needed.) 77.2t ft (b) Find the distance in miles between the two cars after 2 hours 45 minutes. (Round your answer to one decimal place.) 2 mi (c) When are the two cars 1 mile apart? (Round your answer to one decimal place.)

Answer 1

Answer:

Explanation:

The cars are moving away from the intersection at 90° from each other. The motion can be considered a right angled triangle with perpendicular and base being the speeds of the 2 cars. The hypotenuse is the linear distance between them. By Pytagoras theorem:

a) Distance=$\sqrt{a^{2}t + b^{2}t  }$ where a and b are the speeds of 2 cars.

Distance = $\sqrt{40^{2}t + 66^{2}t }$=77.2t feet/second

b) Time = 2 hours 45 minutes; Convert this into seconds to get 9900 seconds. The distance formula will give distance in feet so we will divide it by 5280 to get miles

Distance= (77.2*9900)/5280 = 144.8 miles

c) 1 Miles = 5280 feet.

5280=77.2t;

Time=68.4 seconds;

The 2 cars are 1 miles apart after 68.4 seconds.

Answer 2

The distance between cars can be calculated by the Pythagorean theorem. After 2.45 hrs the distance between cars will be 144.8 miles.

From the Pythagorean theorem

$d = \sqrt {a^2t+ b^2t}$

Where,

$d$- distance

$a$ - speeed of first car = 40 ft/s

$b$ - speed of second car = 66 ft/s

Put the values in the formula,

$d = \sqrt { 40 ^2t+ 66^2t}\\\\d = 77.2 t \rm \ ft/s$

Since the given time = 2.45 hrs = 9900 s

Put the value of $t$

$d = \dfrac {77.2 \times 9900}{5280}\\\\d = 144.8 \rm\ miles$

Therefore, after 2.45 hrs the distance between cars will be 144.8 miles.

Learn more about  the Pythagorean theorem:

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