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    Can someone please help me on these 2 question, like how you work it out and the answer..

    1)A car is travelling along a road at 30ms^-1 When a pedestrian steps onto the road 55m head. The driver of the car applies the brakes after a reaction 0.5s and the car slows down at 10ms^-2. What happens??

    2) The cheetah is the fastest land animal in the world. It can accelerate from rest to 20ms^-1 in 2s, and has a top speed of about 30ms^-1 although it can only maintain this for a distance of 450m before it has to stop and rest. In contrast an antelope can run at around 22 ms^-1 for long periods.

    a) what is the cheetahs acergae acceleration between rest and 20ms^-1
    b) Assume that a cheetah accelerates up to its top speed with the acceleration in ur answer from a, how far will it travel when it accelerates from rest up to its top speed?
    ii) How long does the acceleration take?

    Please help
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    there are several ways of working out the first question one way is
    Work out what distance it would take for a car to stop when starting with a velocity of 30m/s when decelerating at 10m/s using s=ut+1/2 at^2. calculate the distance the car would travel at 30m/s in 0.5s. add this to the distance calculated earlier. If this is greater than 55m/s, the pedestrian's been run over, if not the pedestrian hasn't.
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    1. 0.5s reaction,, so travels 15m before it brakes. then you can use v^2 =u^2 +2as to find how far it travels before stopping. then see if these two are more or less than 55m.

    2. a. use v=u+at and as u is 0 its easy to find a.
    b. i. v^2 =u^2+2as again, sub in the v and a, with u as 0. will give you the s.
    ii. use v=u+at for this one and get your t,
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    (Original post by roar558)
    there are several ways of working out the first question one way is
    Work out what distance it would take for a car to stop when starting with a velocity of 30m/s when decelerating at 10m/s using s=ut+1/2 at^2. calculate the distance the car would travel at 30m/s in 0.5s. add this to the distance calculated earlier. If this is greater than 55m/s, the pedestrian's been run over, if not the pedestrian hasn't.
    Thanks for the help on q1 any chance of 2?
    If not thanks for the help anyway!
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    (Original post by wtf,?)
    1. 0.5s reaction,, so travels 15m before it brakes. then you can use v^2 =u^2 +2as to find how far it travels before stopping. then see if these two are more or less than 55m.

    2. a. use v=u+at and as u is 0 its easy to find a.
    b. i. v^2 =u^2+2as again, sub in the v and a, with u as 0. will give you the s.
    ii. use v=u+at for this one and get your t,
    Thanks for the help and explanation......
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    (Original post by mujahid_e3)
    Thanks for the help on q1 any chance of 2?
    If not thanks for the help anyway!
    2a) use a=(v-u)/t using v=20 u=0 and t=2
    b)use your answer from part a as your acceleration in the equation
    v^2=u^2+2as As u=0, the equation is
    v^2=2as, work out for s
    c)again several ways to do this one, one way is to use s=ut+1/2 at^2 using the a and s values you worked out above.
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    (Original post by wtf,?)
    1. 0.5s reaction,, so travels 15m before it brakes. then you can use v^2 =u^2 +2as to find how far it travels before stopping. then see if these two are more or less than 55m.

    2. a. use v=u+at and as u is 0 its easy to find a.
    b. i. v^2 =u^2+2as again, sub in the v and a, with u as 0. will give you the s.
    ii. use v=u+at for this one and get your t,
    Thanks for the help but how did do u no it travels 15m before it breaks?? in the question it says 55m. and if i use the formula u suggested is it like this: 30^2 +0^2+2*10*55??
    Thanx
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    it is travelling at 30 metres per sec. so if it takes him 0.5 secs to react then it will travel 15 metres before he starts to react and thats where the breaking starts.
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    (Original post by wtf,?)
    it is travelling at 30 metres per sec. so if it takes him 0.5 secs to react then it will travel 15 metres before he starts to react and thats where the breaking starts.
    so what formula do i use and what numbers are in there? Thanks for the help so far!
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    No formula really,, just you get told in one second hel go 30m. so in half a sec you can see easily hel go 15. just 1sec/2 so 30/2.
 
 
 
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