I\'m having trouble with the following question: The Starship Enterprise returns

Question

I'm having trouble with the following question:

The Starship Enterprise returns from warp drive to ordinary spacewith a forward speed of 50 km/h. To the crew's great surprise, aKlingon ship is 100km directly ahead, traveling in the samedirection at a mere 20km/h. Without evasive action, the Enterprisewill overtake and collide with the Klingons in just slightly over3.0 s. The Enterprise's computers react instantly to brake theship. What magnitude acceleration, assuming it's constant, does theEnterprise need to just barely avoid a collision with the Klingonship?

I understand I must equate the position and velocity equations forboth the Enterprise and the Klingon to find the time and thus solvefor the acceleration, but I think I don't have the right equationsfor either, because I just can't seem to get the time. If someonecould please help me out, I would REALLY appreciate it.

Explanation / Answer

. 01.   d = d0 + v0*t +(1/2)*a*t2 02.   v = v0 + a*t 03.   20km/h = (20000 meters/hour)*(1h/3600s) =5.5556 meters/second 04.   50km/h = (50000 meters/hour)*(1h/3600s) =13.8889 meters/second . 05.   dk = distance of klingon = 100000meters + (5.5556 meters/second)*t + (1/2)*(0.0m/s2)*t2 06.   dE = distance of Enterprise = 0.0meters + (13.8889 meters/second)*t + (1/2)*(a)*t2 07.   vE = vk =5.5556 m/s = 13.8889 m/s +a*t      ;   solve fora 08.   a = (-8.3333m/s)/t      ;   substitutethis value for a into '06' 09.   dE = distance of Enterprise = 0.0meters + (13.8889 meters/second)*t + (1/2)*[(-8.3333m/s)/t]*t2 10.   Set '05' equal to '09' to solve for the amountof time between the Klingon and the Enterprise 11.   100000 meters + (5.5556 meters/second)*t =(9.7222meters/second)*t      ;   solvefor t 12.   t = (100000 meters)/(4.1667 meters/second) =24000.0 seconds 13.   a = (-8.3333 m/s)/t = (-8.3333m/s)/(24000.0 seconds) = -347.222E-6m/s2 . 14.   Check: 15.      dk = (100000 m) +(5.5556 m/s)*(24000 s) = 233333.3 meters 16.      dE = 0.0 meters +(13.8889 m/s)*(24000 s) + (1/2)*(-347.222E-6m/s2)*[(24000 s)2] = 233333.3meters 17.      vE = 13.8889 m/s+ (-347.222E-6 m/s2)*(24000 s) = 5.5556 m/s 18.      24000 seconds = 6.667 hours= 6 hours, 40 minutes . 19.   Summary: the Enterprise matches thevelocity of the Klingon ship in 6 hours and 40 minutes at alocation 233,333.3 meters from the Enterprise's originallocation. 20.   Note: I am unsure of the basis forthe 3.0 seconds given in the problem statement. If theKlingon ship was stationary at 100km from the Enterprise, and theEnterprise was traveling at 50km/h, it would require two hours forthe Enterprise to travel a distance of 100km. .

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