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## 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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Physics
3 years
2021-08-28T08:19:57+00:00
2021-08-28T08:19:57+00:00 1 Answers
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## Answers ( )

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