Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Speed and Acceleration Questions-Physics

  • 15-03-2011 9:05pm
    #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: 7,396 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.


Advertisement