Question

Two roads cross at right angles. A police officer sits in a car 65 m east of the crossing and observes a car speeding northbound at 84 m/s. At what speed (in meters per second) is the car distancing itself from the police officer 5 seconds after it passes the crossing?

Answer 1

Answer:83.7496 m/s

Explanation:We can form a right triangle with the given information as follows

Car

|\

420 mt | \

+ | \

84m/s*t | \

| \

crossing |_____\ Police

65mt [t]=seconds

The distance from the police car of the one speeding norhtbound may be calculated using pythagoras:

`d=\sqrt{(420+84*t)^(2)+65^2} , where 420m=84m/s*5 s`

Then the instant speed from the police officer at 5 seconds after the crossing point may be calculated as the derivative with respect to time, evaluated at t=5.

Then first we obtain the derivative.

`(d)/(dt)(√(((420 + 84*t)^2 + 65^2))) =((7056 (t + 5)))/(√((7056 (t + 5)^2 + 4225)))`

Evaluating it when t=5 seconds. We get the speed is 83.7496 m/s, since we remember that speed is displacement over time, and that is what the derivative represents.

Answer 2

Answer: v = 83.012 m/s

Explanation: In a coordinate axis, we may define the x-axis as east and the y-axis as north, and the point where both roads cross can be the (0,0) of our diagram

Then. the police position can be written as (0, 65m)

Here the police see a car speeding at 84 m/s, if we define t= 0 as when the second car passes through the (0,0) point ( the point where the roads cross), the position of the second car can be written as: (84m/s*t, 0).

if we define the distance as a function of time between the police and the other car; we get:

`D(t) = I(65,0) - (84t,0)I = √((84t)^2 + 65^2)`

If we derivate this and valuate it in t = 5, we will obtain the rate at which the distance changes at t = 5 seconds.

`D(t)' = (0.5)/(√((84t)^2 + 65^2) ) *2*84t*84`

and D(5)' = 83.012 m/s

So after 5 seconds, the speed at which the second car is distancing itself from the police car is v = 83.012 m/s