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.)

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= where a and b are the speeds of 2 cars.

Distance = =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.

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

Where,

- distance

- speeed of first car = 40 ft/s

- speed of second car = 66 ft/s

Put the values in the formula,

Since the given time = 2.45 hrs = 9900 s

Put the value of

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

Learn more about  the Pythagorean theorem:

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer