A car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is 28 ft from the intersection. If the speeds of the cars at that instant of time are 9 ft/sec and 13 ft/sec, respectively, find the rate at which the distance between the two cars is changing.

[tex]2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s[/tex]

∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s

batolisis batolisis · Answer 2 · 2022-03-24T16:18:41+01:00

The rate at which the distance between the two cars are changing is :

15.8 ft/s

Given data :

Velocity of car A = 9 ft/s

Velocity of car B = 13 ft/s

Distance car A travels = 21 ft

Distance car B travels = 28 ft

Determine the rate at which the distance between the cars are changing

First step : calculate the distance ( c ) between cars A and B after 4 seconds

applying pythagoras theorem

a² + b² = c² --- ( 1 )

c = √(a² + b² ) = √ ( 21² + 28² ) = √1225 ft

Next step : differentiate with respect to time

2a[tex]\frac{da}{dt}[/tex] + 2b[tex]\frac{db}{dt}[/tex] = 2c [tex]\frac{dc}{dt}[/tex]

[tex]\frac{dc}{dt} = \frac{a\frac{da}{dt}+b\frac{db}{dt} }{c}[/tex] ---- ( 2 )

= [ ( 21 * 9 ) + ( 28 * 13 ) ] / √1225

= 15.8 ft/s

Hence we can conclude that The rate at which the distance between the two cars are changing is : 15.8 ft/s

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