Aman with a very important package needs to get to a post officeimmediately. The

Question

Aman with a very important package needs to get to a post officeimmediately. The nearest post office is 35 miles away, andthis man has 12 minutes to get there before it closes. Theman accelerates according to the following function

                               a(t) = 2t + e^t + 0.5 cos(t)

where theinitial velocity is zero and the position is zero at timezero. Assume t is in minutes and x(t) is in feet.

a) Does hemake it in time?

b) Nowconsider that, after 4 minutes, the man suddenly changes hisacceleration to the following:

             a(v) = 0.04 v^2

Provide the function for his velocitybased on the new acceleration function.

Explanation / Answer

velocity v(t)= a(t)dt=2*t2/2 +et+0.5*sint x(t)= v(t)dt=t3/3 +et-0.5cost x(t)=35 miles=184800 feet with t=12 => x(12)=163330 feet <35 miles=184800 feet=> not in time (b)after t=4 minutes v(4)=70.63 ft/min a(v)=0.04*v2 =dv/dt => 0.04*t= dv/v2 from v=v(4) to v(t) 0.04*t=1/v(4) - 1/v(t) =>v(t)=1/(1/70.63-0.04*t) (b)after t=4 minutes v(4)=70.63 ft/min a(v)=0.04*v2 =dv/dt => 0.04*t= dv/v2 from v=v(4) to v(t) 0.04*t=1/v(4) - 1/v(t) =>v(t)=1/(1/70.63-0.04*t)

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