Question 1010400
With an initial speed of 10.4 m/s,
and a uniform acceleration of 7.50 m/s^2 for 2.00 seconds,
the final speed will be
10.4 + 2.00(7.5) = 10.4 + 15.0 = 25.4 m/s.
The average speed during those 2.00 seconds will be
(1/2)(10.4 m/s + 25.4 m/s)=(1/2)(35.8 m/s) = 17.9 m/s .
With that average speed, in 2.00 seconds, the distance covered would be
2.00(17.9 m/s) = 35.8 meters .
 
WITH FANCY FORMULAS:
An object moving in a straight line during a time {{{t}}} ,
accelerating uniformly with acceleration {{{a}}} ,
from an initial velocity {{{v[0]}}} ,
covers a distance
{{{d=v[0]*t+(1/2)at^2}}} .
In this case, understanding that the units are meters for distance and seconds for time,
{{{system(t="2.00",v=10.4,a="7.50",d=v[0]*t+(1/2)at^2)}}}--->{{{d=10.4*"2.00"+(1/2)("7.50")("2.00")^2=20.8+"15.0"=35.8}}}