Respuesta :
'Acceleration' of 3.1 m/s² means that the taxi is speeding up, and
it's speed is 3.1 m/s faster each second than it was a second earlier.
So, after 6.0 seconds, its speed is (6 x 3.1) = 18.6 m/s faster than
it was at the beginning of the 6 seconds.
That speed was 12.6 m/s when the 6 seconds began, so it's
(12.6 + 18.6) = 31.2 m/s
at the end of the 6 seconds.
Notice that this whole discussion talks about the taxi's speed, not velocity ...
just as the question should. There isn't a single word anywhere in the
question that refers to the taxi's direction, so we don't have a shred of
information that would make it possible to say anything about its velocity.
it's speed is 3.1 m/s faster each second than it was a second earlier.
So, after 6.0 seconds, its speed is (6 x 3.1) = 18.6 m/s faster than
it was at the beginning of the 6 seconds.
That speed was 12.6 m/s when the 6 seconds began, so it's
(12.6 + 18.6) = 31.2 m/s
at the end of the 6 seconds.
Notice that this whole discussion talks about the taxi's speed, not velocity ...
just as the question should. There isn't a single word anywhere in the
question that refers to the taxi's direction, so we don't have a shred of
information that would make it possible to say anything about its velocity.
Answer: 31.2metes/second
Explanation: Correct. When you use the motion equation v = v 0 + at and substitute the values v 0 = +12.6, a = +3.1, and t = 6.0 in the equation, you get the final velocity of the taxi as +31.2 meters/second.
(Edmentum)