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

Help Watch

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    4The Highway Code states that the braking distance for a car travelling at 30 mph (13 m/s) is 14 m. Calculate the deceleration assumed by the Highway Code.

    Didn't do this in class and he gave us this question on kerboodle. Please help me!

    Plus, is decreasing acceleration with velocity as the y axis and time as the x, is decreasing acceleration a curved line ( like the top left quarter of a circle)
    Online

    19
    ReputationRep:
    (Original post by Sraf)
    4The Highway Code states that the braking distance for a car travelling at 30 mph (13 m/s) is 14 m. Calculate the deceleration assumed by the Highway Code.

    Didn't do this in class and he gave us this question on kerboodle. Please help me!

    Plus, is decreasing acceleration with velocity as the y axis and time as the x, is decreasing acceleration a curved line ( like the top left quarter of a circle)
    you can do this with a SUVAT...
    v2=u2+2as
    since the car is decelerating to a halt, the acceleration will be negative and the final velocity is zero.

    ---
    check whether it's velocity v time or distance v time graph - it'd be unusual to deal with non constant acceleration with students who are just beginning.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Joinedup)
    you can do this with a SUVAT...
    v2=u2+2as
    since the car is decelerating to a halt, the acceleration will be negative and the final velocity is zero.

    ---
    check whether it's velocity v time or distance v time graph - it'd be unusual to deal with non constant acceleration with students who are just beginning.
    Exact question is



    4 The Highway Code states that the braking distance for a car travelling at 30 mph (13 m/s) is 14 m. Calculate the deceleration assumed by the Highway Code.

    0.9 m/s 141 m/s 0.5m/s 6m/s
    That's the options
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Joinedup)
    you can do this with a SUVAT...
    v2=u2+2as
    since the car is decelerating to a halt, the acceleration will be negative and the final velocity is zero.

    ---
    check whether it's velocity v time or distance v time graph - it'd be unusual to deal with non constant acceleration with students who are just beginning.
    It's a velocity time graph btw
    Online

    19
    ReputationRep:
    (Original post by Sraf)
    It's a velocity time graph btw
    well if it goes from a steep upward gradient to a less steep upward gradient over a period of time it means that the acceleration has decreased over that period of time.

    are you able to rearrange v2=u2+2as
    to find a?

    v is final velocity - since the car slows to halt this is zero
    u is initial velocity 13 m/s
    s is 14m
    a is the acceleration... and will be a negative number because deceleration is negative acceleration
    Offline

    20
    ReputationRep:
    so if you sketch a velocity-time graph you get a triangle with height 13 and base ?

    the area of the triangle is the distance...

    so 13 x ? /2 = 14

    find ? and then work out the gradient of the hypotenuse.... this is the acceleration.

    the units should be m/s2 not m/s
 
 
 
  • 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
    Will you be richer or poorer than your parents?
  • 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

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