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    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)
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    (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.
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    (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
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    (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
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    (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
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    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
 
 
 
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