Question

1. (a). A police car, approaching right-angled intersection from the north, is chasing a speedmg

Answer 1

Answer:

The SUV is running at 70 km/h

Step-by-step explanation:

Speed As Rate Of Change

The speed can be understood as the rate of change of the distance in time. When the distance increases with time, the speed is positive and vice-versa. The instantaneous rate of change of the distance allows us to find the speed as a function of time.

This is the situation. A police car is 0.6 Km above the intersection and is approaching it at 60 km/h. Since the distance is decreasing, this speed is negative. On the other side, the SUV is 0.8 km east of intersection running from the police. The distance is increasing, so the speed should be positive. The distance traveled by the police car (y) and the distance traveled by the SUV (x) form a right triangle whose hypotenuse is the distance between them (d). We have:

$d=\sqrt{x^2+y^2}$

To find the instant speeds, we need to compute the derivative of d respect to the time (t). Since d,x, and y depend on time, we apply the chain rule as follows:

$\displaystyle d\ '=\frac{x.x'+y.y'}{\sqrt{x^2+y^2}}$

Where x' is the speed of the SUV and y' is the speed of the police car (y'=-60 km/h)

We'll compute  :

$d=\sqrt{(0.8)^2+(0.6)^2}=\sqrt{0.64+0.36}$

$d=1\ km$

We know d'=20 km/h, so we can solve for x' and find the speed of the SUV

$\displaystyle \frac{x.x'+y.y'}{\sqrt{x^2+y^2}}=20$

Thus we have

$x.x'+y.y'=20$

Solving for x'

$\displaystyle x'=\frac{20-y.y'}{x}$

Since y'=-60

$\displaystyle x'=\frac{20+0.6(60)}{0.8}$

$x'=70\ km/h$

The SUV is running at 70 km/h