Question

wo cars start moving from the same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing three hours later?

Answer 1

Answer: Both cars have equal kinetic energy

Explanation:

Answer 2

Answer:

$\frac{dz}{dt} = 65 mi/h$

Explanation:

let distance between two  cars  is = z  mi

we have to find  =$\frac{dz}{dt}$

One travels south at  = 60 mi/h  = $\frac{dx}{dt}$   (given)

the other travels west at =25 mi/h.= $\frac{dy}{dt}$  (given)

since both car have constant speed

at t = 3 hrs

x = 3× 60  = 180 mi/h

y = 3 × 25  = 75 mi/h

from the figure (i)  we get

$z = \sqrt{( x^2+ y^2)}$ ...............(i)

put x and y values

we get

$z = \sqrt{(180)^2 + 75^2}$

$z = \sqrt{32400 + 5625} \\z = \sqrt{38025} \\z = 195 mi/h$

differentiate the equation (i) w r to t

$z^2 = x^2 +y^2$

$2z\frac{dz}{dt} = 2x\frac{dx}{dt}+ 2y\frac{dy}{dt}$

put each values

$2 \times195\frac{dz}{dt} = 2 \times 180\frac{dx}{dt}+2 \times75\frac{dy}{dt}$

$2 \times195\frac{dz}{dt} = 2 \times 180\times 60}+2 \times75\times25\\\frac{dz}{dt} = \frac{{2 \times 180\times 60+2 \times75\times25}}{ 2 \times195}\\\frac{dz}{dt} = 65 mi/h$