A train accelerates at 2.2 m/s^2. With an initial velocity of 22 m/s, what will be the velocity of the train after 19 s

1 answer:

6 0

$a=2.2[\frac{m}{s^2}]\\v_0=22[\frac{m}{s}]\\t=19[s]\\\\a = \frac{\Delta v}{t}\\\Delta v=at=41.8[\frac{m}{s}]\\v = v_0+\Delta v=63.8[\frac{m}{s}]$

Answer: The velocity will be 63.8 m/s.

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