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Let the point that  Car B  started at be  S .

After  t  hours, the distance between  B  and  S  =  70km/hour * t hours  =  70t km

Affter  t  hours, the distance between  A  and  S  =  40 km - 40km/hour * t hours  =  (40 - 40t) km

So after  t  hours, the distance between  A  and  B   =   √[  (70t)²  +  (40 - 40t)²  ]

distance between  A  and  B   =   √[  4900t²  +  1600 - 3200t + 1600t²  ]

distance between  A  and  B   =   √[  6500t² - 3200t + 1600  ]

distance between  A  and  B   =   √[  100(65t² - 32t + 16)  ]

distance between  A  and  B   =   10√[  65t² - 32t + 16  ]

Let the distance between  A  and  B  be  y .

y   =   10√[  65t² - 32t + 16  ]

We want to find what value of  t  produces the minimum value of  y  in that equation.

dy/dt   =   d/dt( 10√[  65t² - 32t + 16  ]  )

dy/dt   =   10 d/dt √[  65t² - 32t + 16  ]

dy/dt   =   10 (1/2) ( 65t² - 32t + 16 )^(-1/2) d/dt( 65t² - 32t + 16 )

dy/dt   =   10 (1/2) ( 65t² - 32t + 16 )^(-1/2) (130t - 32)

Set  dy/dt  equal to zero.

0   =   10 (1/2) ( 65t² - 32t + 16 )^(-1/2) (130t - 32)

Divide both sides by 10, multiply both sides by 2, multiply both sides by  ( 65t² - 32t + 16 )^(1/2)

0   =  (130t - 32)

32  =  130t

t  =  16/65       hours

This is the only value of  t  that makes  dy/dt  be  0, so this is the value of  t  that minimizes  y .

When  t  =  16/65 ...

y   =   10√[  65(16/65)² - 32(16/65) + 16  ]

y   =   56√[65] / 13     kilometers