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A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt.The acceleration during the first5.6 km of travel is closest to:
a) 0.19 m/s2
b) 0.14 m/s2
c) 0.16 m/s2
d) 0.20 m/s2
e) 0.17 m/s2

Respuesta :

Answer:

c) 0.16 m/s2

Explanation:

The computation of the acceleration during the first km of travel is shown below

Given that

Final velocity = v = 42 m/s

Initial velocity = u = 0 m/s

Distance = 5.6km

Based on the above information, we need to apply the following formula

As we know that

[tex]v^2 - u^2 = 2as[/tex]

[tex]a = \frac{v^2 - u^2}{2s}[/tex]

[tex]= \dfrac{42^2 - 0^2}{2 \times 5.6 \times \frac{1000\ m}{1\ km} }[/tex]

= 0.1575 m/s ^2

hence, the correct option is c. 0.16 m.s^2

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