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Speed and Acceleration Questions-Physics

  • 15-03-2011 10:05pm
    #1
    Chuchoter
    Registered Users, Registered Users 2 Posts: 4,305 ✭✭✭


    About to lose patience with these, how do you do the questions where its like a bike sets off a 4ms^-1 and 4 seconds later a car starts off at 5ms^1, when and where do they meet? Or the ones where its rocks falling of cliffs meeting one getting thrown up towards it. I've been trying to do the Section F questions of the Workbook for Real Physics of Acceleration and not getting anywhere. Whats the trick?


Comments

  • #2
    **Timbuk2**
    Moderators, Education Moderators, Motoring & Transport Moderators Posts: 7,395 Mod ✭✭✭✭**Timbuk2**


    For the bike/car one, you use the s=ut+.5at^2 formula to get the distance travelled after t seconds (if they are going constant velocity, then distance=speedxtime, or use the above and set a=0).

    They will meet when the displacements are equal. Solve for t. You now know when, and can just sub in the t to get where.

    It's roughly the same idea for the rocks falling off the cliffs - BUT, remember that acceleration is caused/hindered by gravity, so a=+/- 9.8 depending on which direction it is thrown. Also, s is displacement, and not distance travelled, so if it happens that the rock that is thrown up reaches a max height and starts to fall again before it meets the other rock, then you need to be careful with these types of questions. For getting it to meet a falling rock, then it doesn't matter, but in some circumstances it does - remember at a max height, v=0.


  • #3
    Chuchoter
    Registered Users, Registered Users 2 Posts: 4,305 ✭✭✭Chuchoter


    I do that and it never works out :(I'm always out by like 1.


  • #4
    RHRN
    Closed Accounts Posts: 357 ✭✭RHRN


    Go like this.

    Use s1=Bike distance
    s2 = Car distance

    now set s1=s2

    Using s=(ut)(1/2.at^2)
    Which in this case is simply s=ut

    We have s1=s2
    4(t)=5(t-4)

    Remember, the car is travelling for four seconds less than the bike!

    In this case, we solve for t=20.
    Then put 't' into s1 to get 80m as our where.

    As for rocks, it generally works the same way. Remember to put in 'g' for a, -g that is. Where one rock is falling, its actually okay as for example, whenever you solve for t in an object thrown upwards, you'll get a quadratic with two solutions as the object is there on the way up and on the way down, so setting s1=s2, with -g=a, and using the correct t's (as shown above) will work.


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