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Solve for the required. 1.)Avan is 5 km due north in the afternoon. if the van is travelling northward at the rate of 60 kph

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Solution:

Since it is not given very clearly, i assume that the van is 5km due north in the beginning

and at that time the bus is at the origin

So the distance between the van and bus at 2:pm would be

So the distance between the van and bus any time after 2 pm would be

d(t)ー (5 + 60が (75t)2

dd(t) 60 (5 60t) 752t dt (5 +60t)+(75t)

dd(t) 300 9225t dt 5 +60t)(75t)-

dd(t) 300 + 9225 * 3 (5+60-3)2+ (75.3)

dd(t) 300 + 9225 * 3 t-3- dt (185)2 + (225)2

\frac{\mathrm{d} d(t)}{\mathrm{d} t}|_{t=3}=\frac{27975}{\sqrt{84850}}

\frac{\mathrm{d} d(t)}{\mathrm{d} t}|_{t=3}\approx 96

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