Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    i have this question that keeps bugging me:
    i know that suvat equations apply to questions with constant acceleration, but does this include when a = 0?
    Offline

    15
    ReputationRep:
    (Original post by blankboi)
    i have this question that keeps bugging me:
    i know that suvat equations apply to questions with constant acceleration, but does this include when a = 0?
    Yes, the suvat equations work for a = 0, since this is a constant value for a.


    When a = 0:

    v = u + at --> v = u, which makes sense if there is no acceleration. Same for v^2 = u^2 + 2as.

    s = ut + 0.5*at^2 --> s = ut, which is the usual equation for zero acceleration (distance = speed x time).

    s = t(u+v)/2 --> s = t(2u)/2 = ut, as above.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by ombtom)
    Yes, the suvat equations work for a = 0, since this is a constant value for a. All of the equations should reduce to distance = speed x time.
    thanks

    but this makes me confused after doing this question:
    A steel ball of mass 0.15kg, released from rest in a liquid, falls a distance of 0.2m in 5 seconds. Assuming the ball reaches terminal velocity within a fractio of a second, calculate:
    i) its terminal speed

    I originally used the suvat equations to get speed but i got 0. 08 (answer is 0.04) and im not sure why its wrong since a would be 0.

    i know that i need to use the speed = distance / time equation, but this is for constant velocity and that means 0 acceleration.
    Offline

    15
    ReputationRep:
    (Original post by blankboi)
    thanks

    but this makes me confused after doing this question:
    A steel ball of mass 0.15kg, released from rest in a liquid, falls a distance of 0.2m in 5 seconds. Assuming the ball reaches terminal velocity within a fractio of a second, calculate:
    i) its terminal speed

    I originally used the suvat equations to get speed but i got 0. 08 (answer is 0.04) and im not sure why its wrong since a would be 0.

    i know that i need to use the speed = distance / time equation, but this is for constant velocity and that means 0 acceleration.
    speed = distance/time = 0.2/5 = 0.04 m/s. What were you trying to do?
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by ombtom)
    speed = distance / time = 0.2/5 = 0.04 m/s. What were you trying to do?
    I first did s = (u+v)t/2
    and so rearranged to (2s/t) = v and this equals to 0.08 which is wrong

    when using the speed = distance / time equation, does that mean a = 0?
    if it does, then i don't understand why it doesn't work for suvat
    Offline

    15
    ReputationRep:
    (Original post by blankboi)
    I first did s = (u+v)t/2
    and so rearranged to (2s/t) = v and this equals to 0.08 which is wrong

    when using the speed = distance / time equation, does that mean a = 0?
    if it does, then i don't understand why it doesn't work for suvat
    Speed = distance/time only applies when a = 0.

    Suvat equations only apply when a = constant (0 is a constant).

    Variable acceleration requires differentiation/integration.

    s = (u+v)t/2.

    As a = 0, u = v, so s = 2ut/2 = ut.

    Your mistake was ignoring u; u = v, so u + v = 2u, or 2v.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by ombtom)
    Speed = distance/time only applies when a = 0.

    Suvat equations only apply when a = constant (0 is a constant).

    Variable acceleration requires differentiation/integration.

    s = (u+v)t/2.

    As a = 0, u = v, so s = 2ut/2 = ut.

    Your mistake was ignoring u; u = v, so u + v = 2u, or 2v.
    hm i get what your'e saying but it says it was released from rest so u = 0?
    Offline

    15
    ReputationRep:
    (Original post by blankboi)
    hm i get what your'e saying but it says it was released from rest so u = 0?
    It says it accelerates to terminal velocity in a negligible amount of time. After it has reached terminal velocity, a = 0. As it had to accelerate for a short period of time, the answer is an approximation.

    We're applying v = d/t from the time when it reaches its maximum velocity.

    You're right that the speed starts as 0, but it accelerates quickly; so quickly, that we can consider u to be a positive value.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.