 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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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) $$0.16\frac{m}{s}$$