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Dafna1 [17]
4 years ago
6

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 const

ant 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.?

Mathematics
2 answers:
Charra [1.4K]4 years ago
8 0

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.

Rufina [12.5K]4 years ago
3 0

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

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