Question

Two cars are traveling along perpendicular roads, car A at 40 mi/hr, car B at 60 mi/hr. At noon, when car A reaches the intersection, car B is 90 mi away, and moving toward it. At 1 p.m. the distance between the cars is changing, in miles per hour, at the rate of:

Answer 1

Answer:

$\frac{dD}{dt} = -4 miles/hour$

negative sign indicates that the distance is decreasing with time

Explanation:

Let at any time t after noon that is 12 p.m.  

distance traveled by car A = 40t

distance traveled by car B = 90-60t

then distance between the two cars at time t

$D^2= (40t)^2+(90-60t)^2$............1

also, at time 1 p.m.

distance $D^2= (40\times1)^2+(90-60\times1)^2$

D=50 Km

differentiating equation 1 w.r.t. t we get

$2D\frac{dD}{dt}= 2\times40t\times40+2(90-60t)(-60)$

put t= 1 and D= 50 we get

$2\times50\frac{dD}{dt}= 3200\times1-3600\times1$

$\frac{dD}{dt} = -4 miles/hour$