An ice sled powered by a rocket engine starts from rest on a large frozen lake and ac

  1. An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +38 ft/s2. After some time t1, the rocket engine is shut down and the sled moves with constant velocity v for a time t2. Assume the total distance traveled by the sled is 15750 ft and the total time is 90 s.
    (a) Find the times t1 and t2.
    t1
    t2
    (b) Find the velocity v.
    ft/s
    At the 15750 ft mark, the sled begins to accelerate at -21 ft/s2.
    (c) What is the final position of the sled when it comes to rest?
    ft
    (d) What is the duration of the entire trip?

    Please just show me how to do this. My entire weekend has consisted of me holed up in my basement doing physics. This is my last problem and I just want to feel happy and alive again :(
    s
     
  2. jcsd
  3. cepheid

    cepheid 5,190
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: An ice sled powered by a rocket engine starts from rest on a large frozen lake an

    Welcome to PF funwithphysics!

    No, please read our site rules. We won't do your homework for you here. Instead, let's go through this step by step. Start with the first phase of the motion, between t = 0 and t = t1. This is the constant acceleration portion of the trip. Write down an expression for the distance vs. time during this phase.

    Next, write an expression for the distance vs. time during the second portion, between t = t1 and t = t2. This is a constant velocity portion of the trip, and you know what v is in terms of 'a' and t1.

    You know that the two distances have to add up to 15,750 ft, and that the two times have to add up to 90 s.

    That gives you two equations, and two unknowns. So you can solve for both of them.
     
  4. Re: An ice sled powered by a rocket engine starts from rest on a large frozen lake an

    OK . Your last line was funny.

    From your problem
    Eq 1 : t1 + t2 = 90 seconds

    also, distance travelled during acceleration + distance at constant velocity v is 15750 feet
    Eq 2: d1 + d2 = 15750 feet

    acceleration,
    Eq 3: a1 = 38 ft/s^2
    Eq 4: a2 = 0



    From an equation of linear motion,
    x = x0 + 1/2 vt
    x = x0 + v0 t + 1/2 a t^2
    v = v0 + at
    v^2 = v0^2 + 2 a d

    Can you think how to manipulate those equations to solve for t1, t2 and v in your problem?
     
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