Hey there Sign in to join this conversationNew here? Join for free

suvat

Announcements Posted on
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    a couple of question involving Suvat.....

    "1. A car, moving with uniform acceleration along a straight level road, passed points A and B when moving with speed 30ms^{-1} and 60ms^{-1} respectively. Find the speed of the car at the instant it passed C, the mid-point of AB."

    i thought C, being midpoint you could just find the average of the two velocitys given.

    so, (30+60)/2 = 45ms^-1

    however, the answer is 47.4ms^-1, i do have an assumption that it is a typo though, just need comfirmation.

    "2. A car is moving along a straight horizontal road at a constant speed 18ms^{-1}. at the instant when the car passed a lay-by, a motor-cyclist leaves the lay-by, starting from rest, and moves with constant acceleration 2.5 ms^{-2} in pursuit of the car. Given that the motor-cyclist overtakes the car T seconds after leaving the lay-by, calculate:

    a, the value of T
    b, the speed of the motor-cyclist at the instant of passing the car."

    i really dont understand how this can be drawn out..mainly because there are 2 vehicles and start-stop, is this suppose to be modelled using a speed-time graph?
    • 29 followers
    Offline

    For part 1. the book is correct.

    If the velocity were increasing uniformily and you were asked to find the velocity when it had been travelling for half the time you would be correct.

    However, it is asking the velocity at half the distance.

    See spoiler for how to do first part.

    Spoiler:
    Show

    Use V^2=u^2+2as for the whole journey A to B.
    Assuming it start at A. This will give you "as".
    Repeat the process for the half way point, substituting in your value of "as".

    • 0 followers
    Offline

    ReputationRep:
    (Original post by KeineHeldenMehr)
    i really dont understand how this can be drawn out..
    2. a) For motor-cyclist, u=0 a=2.5 t=T s=?. For car, s=18xT. Use simultaneous equations.
    2. b) You will have t so find v for motor-cyclist.
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    (Original post by ghostwalker)
    For part 1. the book is correct.

    If the velocity were increasing uniformily and you were asked to find the velocity when it had been travelling for half the time you would be correct.

    However, it is asking the velocity at half the distance.

    See spoiler for how to do first part.

    Spoiler:
    Show

    Use V^2=u^2+2as for the whole journey A to B.
    Assuming it start at A. This will give you "as".
    Repeat the process for the half way point, substituting in your value of "as".

    s=?
    u30
    v=60
    a=?
    t=?

    v^{2}=u^{2}+2as

    60^{2}=3=^{2}+2as

    2as=2700

    as=1350

    hmm...
    Repeat the process for the half way point, substituting in your value of "as"
    v^{2}=u^{2}+2(1350)

    yeah, i'm stumped.
    • 29 followers
    Offline

    For the second part the distance is half AB.
    So, as/2

    or more fully v^2=u^2+2\frac{as}{2}
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    (Original post by ghostwalker)
    For the second part the distance is half AB.
    So, as/2
    ah gotcha~

    v= \sqrt{30^{2}+1350}

    =47.4....ms^-1.

    thank you.
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    (Original post by OllyThePhilosopher)
    2. a) For motor-cyclist, u=0 a=2.5 t=T s=?. For car, s=18xT. Use simultaneous equations.
    2. b) You will have t so find v for motor-cyclist.
    that worked wonders! thank you!
    • 0 followers
    Offline

    ReputationRep:
    Generally you need three unknowns for this to work. In the first one it works because you are halving an initial distance. For the second part you know that t and s are the same for both. When they meet, if time 0 is when the car passes, they will have traveled the same distance and time.

    Work out s in terms of t for motorbike. For the car just use v=s/t

    See if you can get it from there.

    Although generally, these seem harder than the typical exam questions I got.
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    ^thanks.

    another question which i think i have solved however, little understanding to what is going on.
    "A particle P moved in a straight line with constant retardation. At the instants when P passed through the points A, B and C it was moving with speeds: 10ms^{-1}, 7ms^{-1}, 3ms^{-1} respectively.

    prove that: \frac{AB}{BC}=\frac{51}{40}.

    going back to the first question in this thread, i use the same method "ghostwalker" has suggested....working out in terms of "as" for AB and then BC.

    AB:
    s=?, u=10,v=7, a=?, t=?

    v^{2}=u^{2}+2as

    as=\frac{-51}{2} (-25.5)

    BC:
    s=?, u=7,v=3,a=?,t=?

    v^{2}=u^{2}+2as

    as=\frac{-40}{2}

    so, \frac{-51}{2} \div \frac{-40}{2}

    =\frac{AB}{BC}=\frac{51}{40}.

    im just unsure why this can be done in terms of 'as'
    surely as/as equals 1, so,

    i know i dont make sense, but this is because i cant make sense out of this:

    \frac{AB}{BC}=\frac{as}{as}=\fra  c{\frac{-51}{2}}{\frac{-40}{2}}

    1=\frac{51}{40}.

    what happens to the 1?
    • 29 followers
    Offline

    You're using different values of s each time.
    To clarify call the distance AB "b", and BC "c", then you will get values for "ab" and "ac".
    When you divide you get b/c = (AB)/(BC)
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    (Original post by ghostwalker)
    You're using different values of s each time.
    To clarify call the distance AB "b", and BC "c", then you will get values for "ab" and "ac".
    When you divide you get b/c = (AB)/(BC)
    why did you clarify in that way though?
    :confused:
    • 29 followers
    Offline

    Not sure what you mean by that question.

    The letter s represents two different things in your workings, so dividing "as" where s means the distanc AB by "as" where s means the distance BC is what I assumed you saw as confusing.

    Hence my suggestion to use b, and c.
    If you call the first distance b, and the second one c, you don't end up with (as)/(as).

    They don't have to be b and c, just any two letters.
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    (Original post by ghostwalker)
    Not sure what you mean by that question.

    The letter s represents two different things in your workings, so dividing "as" where s means the distanc AB by "as" where s means the distance BC is what I assumed you saw as confusing.

    Hence my suggestion to use b, and c.
    If you call the first distance b, and the second one c, you don't end up with (as)/(as).

    They don't have to be b and c, just any two letters.
    ok, thanks....
    • 29 followers
    Offline

    Alternatively, when you get the result of the first part you can divide through by a, and say distance AB = -25.5/a.
    Similarly distance BC = -20/a.
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    (Original post by ghostwalker)
    Alternatively, when you get the result of the first part you can divide through by a, and say distance AB = -25.5/a.
    Similarly distance BC = -20/a.
    yeah that works,

    s/s, a/a = 1

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

Updated: December 23, 2008
New on TSR

Moving on from GCSEs

What advice would you give someone starting A-levels?

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.