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    is this M3?
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    The question was solved using an assumption:
    1. The car was at rest at the bottom of the hill.

    Since the car is moving uphill, a component of gravity will be acting it, in downwards direction. Let this be g' (where g' = g sin 15).

    Now, using s = ut + (1/2) a t2, where the symbols have their usual meaning,
    distance travelled in 30 sec :

    s = (1/2)(a-g')(302) = 450 (a-g').

    after switching off the engine, the car comes to rest at some point. now, v = 0.

    use v2 - u2 = 2 a s.
    note that:
    v=0
    u = (a-g')(30) (using v=u+at)
    a= (-g') (since it is in opposite direction of initial motion).

    so, distance travelled after switching off the engine:
    s' = [ { 30(a-g') }2 ] / (2 g')

    s' = 450 ( (a2)/g' - 2a + g' )

    so, the 2 distances measured can be equal only if (a2)/g' - 2a + g' = (a2)/g'

    so, 2a - g'= 0.

    therefore, a = (1/2) g sin 15.

    now, taking g= 9.8 and sin 15 = 0.25,

    a = 1.225.

    but a is given as 2.5, which means g'-2a is negative, so the car moved more in the first 30 seconds than for the rest of it's journey upwards.

    Hope this helps.
 
 
 
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