Question

A car traveling east at a constant rate of 33 mi/hr passes through an intersection at 10 A.M. A truck traveling north at a constant rate of 42 mi/hr passes through the same intersection at 11 A.M. If both vehicles maintain their speed and direction, how fast is the distance between them increasing at 1 P.M.?

Answer 1

Answer:52.33 m/s

Step-by-step explanation:

Given

Speed of car $v_1=33 mi/hr$

Speed of truck $v_2=42 mi/hr$

Distance traveled by car in 1 hr is 33 mi

Therefore after 11 am distance of car and truck from Junction in time t is  

Car $=33+33t$

truck $=42 t$

Let D be the distance between them

$D^2=42t^2+(33+33t)^2$------------1

at $t= 2 hr$

$D^2=(42\times 2)^2+(33+66)^2$

$D=129.83 miles$

To get how fast distance between them is increasing differentiate 1 we get

$2D\frac{\mathrm{d} D}{\mathrm{d} t}=2\cdot(42)^2t+2\cdot 33^2(1+t)$

$\frac{\mathrm{d} D}{\mathrm{d} t}=\frac{42^2t+33^2\left ( 1+t\right )}{D}$

$\frac{\mathrm{d} D}{\mathrm{d} t}=\frac{1764\times 2+1089\times 3}{D}$

$\frac{\mathrm{d} D}{\mathrm{d} t}=\frac{6795}{129.83}=52.33 m/s$

Answer 2

Answer:

52.34 miles per hour

Step-by-step explanation:

Let x represents the distance covered by car and y represents the distance covered by truck,

Also, suppose l represents the distance between them,

∵ car is travelling in east direction while truck is travelling in north direction,

So, by the Pythagoras theorem,

$l^2 = x^2 + y^2----(1)$

Differentiating with respect to t ( time ),

$2l \frac{dl}{dt}=2x\frac{dx}{dt}+3y\frac{dy}{dt}$

$\implies l \frac{dl}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$,

Now, the speed of car is 33 mi/hr and speed of truck is 42 mi/hr,

i.e $\frac{dx}{dt}=33\text{ mi per hour}\text{ and }\frac{dy}{dt}=42\text{ mi per hour}$

$\implies l \frac{dl}{dt}=33x+42y-----(2)$,

Distance = speed × time,

So, y = 42 × 2 = 84 miles,

x = 33 × 3 = 99 miles ( ∵ car travelled 3 hours till 1 PM while truck travelled 2 hours till 1 PM)

From equation (1),

$l = \sqrt{84^2 + 99^2}=\sqrt{7056+9801}=\sqrt{16857}=129.83$

From equation (2),

$129.83\frac{dl}{dt}=33(99)+42(84) = 3267+3528=6795$

$\implies \frac{dl}{dt}=\frac{6795}{129.83}=52.34\text{ miles per hour}$

Hence, the distance between them is increasing by 52.34 miles per hour.