A train starts from rest and accelerates uniformly, until ithas traveled 5.6 km and acquired a velocity of 42 m/s. Thetrain then moves at a constant velocity of 42 m/s for 420 s.

glasskerfu 2021-02-21 Answered

A train starts from rest and accelerates uniformly, until ithas traveled 5.6 km and acquired a velocity of \(42 \frac{m}{s}\). Thetrain then moves at a constant velocity of \(42 \frac{m}{s}\) for \(420 s\). The train then slows down uniformly at \(0.065 \frac{m}{s^{2}}\), untilit is brought to a hault. The acceleration during the first \(5.6 km\) of travel is closest to:
a) \(0.19 \frac{m}{s^{2}}\)
b) \(0.14 \frac{m}{s^{2}}\)
c) \(0.16 \frac{m}{s^{2}}\)
d) \(0.20 \frac{m}{s^{2}}\)
e) \(0.17 \frac{m}{s^{2}}\)

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Expert Answer

hosentak
Answered 2021-02-22 Author has 27853 answers

here is a problem with a train in motion & the accfor the 1st 5.6km is asked so you don't have to think of the later part.concentrate on the givendata:--
1. initial velocity\(=0\frac{m}{s}(u)\)
2. final velocity\(=42\frac{m}{s}(v)\)
3. distance\(=5.6km=5600m(s)\)
so put it in the formula--\(v2=u2+2(a)\)
(s) \(\&\) the ans will be (c) \(0.16\frac{m}{s}\)

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