Advertisement
Help Keep Boards Alive. Support us by going ad free today. See here: https://subscriptions.boards.ie/.
If we do not hit our goal we will be forced to close the site.

Current status: https://keepboardsalive.com/

Annual subs are best for most impact. If you are still undecided on going Ad Free - you can also donate using the Paypal Donate option. All contribution helps. Thank you.
https://www.boards.ie/group/1878-subscribers-forum

Private Group for paid up members of Boards.ie. Join the club.

kinematics

  • 17-05-2010 02:43PM
    #1
    Registered Users, Registered Users 2 Posts: 35


    can someone please help with this question... i am really stuck on it....

    a cyclist travelling at 8m/s obvserves a car, 40m distant, just beginning to accelerate away uniformly from rest. if the cyclists speed remains the same he can just catch the car.

    (i) find the acceleration of the car.

    (ii) if he had been cycling at 6m/s how close would he have got to the car?


    any help would be appreciated. cheers. :)


Comments

  • Closed Accounts Posts: 104 ✭✭MariaBabii


    (i)
    v2 = u2 + 2as
    v - speed - 8
    u - rest - 0
    a - acceleration - ?
    s - distance

    (8)2 = 0 + 2a(40)
    64 = 80a
    a = 0.8
    0.8 x 60 = 48m/s

    (ii)(6)2 = 0 + 2a(40)
    36 = 80a
    a = 0.45
    0.45 x 60 = 27m/s

    Hahaha just a guess ... i havent done applied maths in 2years lol


  • Registered Users, Registered Users 2 Posts: 581 ✭✭✭Ruski


    MariaBabii wrote: »
    (i)
    v2 = u2 + 2as
    v - speed - 8
    u - rest - 0
    a - acceleration - ?
    s - distance

    (8)2 = 0 + 2a(40)
    64 = 80a
    a = 0.8
    0.8 x 60 = 48m/s

    (ii)(6)2 = 0 + 2a(40)
    36 = 80a
    a = 0.45
    0.45 x 60 = 27m/s

    Hahaha just a guess ... i havent done applied maths in 2years lol

    That's not right, since the cyclist travels with uniform acceleration, and doesn't stop.

    I'll have an answer out for you in ten minutes.


  • Registered Users, Registered Users 2 Posts: 581 ✭✭✭Ruski


    Cyclist:
    u = 8m/s
    v = 8m/s
    a = 0m/s^2 (Since he travels at constant speed)
    s = 40m
    t = ?

    (i)
    Find t first:
    s=(u)(t)+1/2(a)(t)^2
    40=(8)(t) [since a=0]
    t=5s


    Use distance formula again this time for car, whose acceleration is [a]
    s=ut+1/2(at)^2
    40=1/2(a)(5)^2 [when u=0, since car starts at rest]
    40=25/2 (a)
    a=80/25
    a=3.2m/s^2

    (ii)


    How close:
    s = ut
    s = (6)(5)
    s = 30m

    He gets 10m close to the car since 40 - 30 is 10.


  • Registered Users, Registered Users 2 Posts: 35 rushrovers


    Ruski wrote: »
    Cyclist:
    u = 8m/s
    v = 8m/s
    a = 0m/s^2 (Since he travels at constant speed)
    s = 40m
    t = ?

    (i)
    Find t first:
    s=(u)(t)+1/2(a)(t)^2
    40=(8)(t) [since a=0]
    t=5s


    Use distance formula again this time for car, whose acceleration is [a]
    s=ut+1/2(at)^2
    40=1/2(a)(5)^2 [when u=0, since car starts at rest]
    40=25/2 (a)
    a=80/25
    a=3.2m/s^2

    (ii)


    How close:
    s = ut
    s = (6)(5)
    s = 30m

    He gets 10m close to the car since 40 - 30 is 10.

    thank you so much thats great! thanks to both of yous.....


Advertisement