A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 52.0 km/h in the 16.0 s that it takes to round the bend. Th

Question

A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 52.0 km/h in the 16.0 s that it takes to round the bend. The radius of the curve is 120 m. Compute the acceleration at the moment the train speed reaches 52.0 km/h. Assume the train continues to slow down at this time at the same rate.

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Latifah 2 months 2021-08-28T08:19:57+00:00 1 Answers 0 views 0

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    2021-08-28T08:21:15+00:00

    Answer:

    Explanation:

    Given a circular motion

    A train slows down to as it want to round the curve from

    u = 86km/hr = 86×1000/3600=23.89m/s

    v = 52km/hr= 52×1000/3600=14.44m/s

    Time taken is 16s

    Radius of curve is 120m

    Now, tangential acceleration a(t)

    a(t) = dV/t

    a(t) =(v-u)/t

    a(t) =(14.44-23.89)/16

    a(t) =-9.444/16

    a(t) =-0.59m/s²

    So, now the centripetal acceleration a(c) is given as

    a(c) =v²/r

    a(c) =14.444²/120

    a(c) =208.64/120

    a(c) =1.74m/s²

    Now, total acceleration is

    a=√(a(t)²+a(c)²)

    a=√((-0.59)²+1.74²)

    a=√(0.3481+3.025)

    a=√3.3731

    a=1.84m/s²

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