Question

A spaceship leaves the earth, starting from rest and accelerating at a constance rate of 10m/s^2. its destination is the Moon, 380,000 km from earth. ignore size of earth and moon. Assume that both are at rest.

a.) At this rate of acceleration how long will it take the spacecraft to reach the moon?

for part b assume the spaceship accelerates to a halfway point (same acceleration as a)and then begins to deccelerate to have a final velocity of 0 when it reaches the moon.

b.) With the new method how long will it take for the spaceship to reach the moon?

the spaceship is carrying a probe which it can launch ouot of the front end of the spacecraft with a speed of 5000 m/s with respect to the spaceship.

c.) Assume the spaceship is traveling the same way as part b, at what distance from the moon should the probe be launched so that it will reach the moon at the same time as the saceship?

suppose the spaceship also has a probe that it can launch in the backward direction at the same speed of 5000 m/s with respect to the ship.

d.) In this case what distance from the moon should the probe be launched so that it will reach the moon at the same time as the ship?

Need to know how to do this type of work for my exam. An explained answer so i can follow along is greatly appreciated thank you in advance.

Answer 1

We are given the acceleration ---    \(a=10 meters/sec^{2}\) , and the distance from the earth to moon is ---    \(D=380,000km=380,000,000meters\) .

(a) For constant accelerated motion from rest , we then need :

   \(D=380,000,000=1/2at^{2}=1/2(10)t^{2}=5t^{2}\) ===>

   \(t^{2}=380,000,000/5=76,000,000\) ; so , the time is :

\(t=\sqrt{76,000,000}=2,000\sqrt{19}\approx 8,717.798seconds\) or    \(2hours25minutes17.798seconds\) ---- SOLUTION .

(b) For the first half of D ---    \(1/2D=190,000,000meters\) , we have constant accelerated motion at    \(10meters/sec^{2}\) :

   \(1/2D=190,000,000=1/2(10)t_{1}^{2}=5t_{1}^{2}\) ===>

   \(t_{1}^{2}=190,000,000/5=38,000,000\) ===>

   \(t_{1}=\sqrt{38,000,000}=1000\sqrt{38} \approx 6,164.41400297seconds\) .

The velocity of the ship after t1 seconds is :

   \(v(t_{1})=at_{1}=10t_{1}=61,644.1400297meters/sec\) . For the second half trip of 1/2D , we have an acceleration of    \(-a=-10meters/sec^{2}\) . We then have velocity :

   \(v=dx/dt=-10t+61,644.1400297\) . Integrate both sides , and we get :

   \(x=-5t^{2}+61,644.1400297t\) . We then need this value of x to equal 190,000 ,000:

   \(x=-5t_{2}^{2}+61,644.1400297t_{2}=190,000,000\)   ===>

   \(5t_{2}^{2}-61,644.1400297t_{2}+190,000,000=0\)   . Use the quadratic formula to solve for the smaller value of

   \(t_{2}\)   :

   \(t_{2}=(61,644.1400297-\sqrt{(-61,644.1400297)^{2}-4(5)190,000,000})/10 \approx 6,164.41400297seconds\)   

We then have the total time for the trip is :

   \(t=t_{1}+t_{2} \approx 12,328.8280059seconds\)   or    \(3hours25minutes28.828seconds\)   --- SOLUTION.

For parts (c) and (d) , the probe must be launched during the first half trip while the ship is accelerating because otherwise the ship will never catch up to probe .

(c) Call the time when the probe is launched during the first half trip    \(t_{o}\)   . The velocity of the ship is then ---    \(v_{ship}=at_{o}=10t_{o}\)   , and the constant velocity of the probe is ---    \(v_{p}=v_{ship}+5000=10t_{o}+5000\)   . The time left for the probe to reach the moon is :

   \(t_{left}=12,328.8280059-t_{o}\)   . At t=to , the ship has traveled :

   \(x=5t_{o}^{2}\)   . The distance remaining for the probe to reach the moon is then :

   \(D_{left}=380,000,000-5t_{o}^{2}\)   . The probe travels at constant speed    \(v_{p}\)   ; so , we must have :

   \(D_{left}=380,000-5t_{o}^{2}=v_{p}t_{left}=(10t_{o}+5000)(12,328.8280059-t_{o})\)   . Multiply the quantities on the right , and we get :

   \(380,000,000-5t_{o}^{2}=-10t_{o}^{2}+118,288.280059t_{o}+61,644,140.0295\) .Simplifying , we get the quadratic equation :

   \(5t_{o}^{2}-118,288.280059t_{o}+318,355,859.970=0\)   . Use the quadratic frormula to solve for the smaller value of to :

   \(t_{o}=(118,288.280059-\sqrt{(-118,288.280059)^{2}-4(5)(318,355,859.97)})/10 \approx 3,096.703seconds\)   

The probe is then launched when the distance of the ship from the moon is :

   \(D_{left}=380,000,000-5t_{o}^{2} \approx 332,052,140meters\)   --- SOLUTION .

(d) Call the time again to when the probe is launced during the first half trip . The only thing that changes is the velocity of the probe ---       \(v_{p}=10t_{o}-5000\)   . We then solve the equation :

   \(D_{left}=380,000,000-5t_{o}^{2}=v_{p}t_{left}=(10t_{o}-5000)(12,328.828-t_{o})\)   which simplifies to the quadratic equation :

   \(5t_{o}^{2}-128,288.280059t_{o}+441,644,140.030=0\)   . Using the quadratic formula to solve for the smaller value of to , we get :

   \(t_{o}=(128,288.280059-\sqrt{(-128,288.280059)^{2}-4(5)441,644,140.03})/10 \approx 4,096.703seconds\)

The probe is then launched when the distance of the ship from the moon is :

     --- SOLUTION .