If a particle accelerates from #"6.9 m/s"# at a rate of #"0.62 m/s"^2# over a time span of #"3.4 s"#, what is the new speed?
1 Answer
Oct 22, 2017
Well, you should expect it to be larger.
Given an acceleration is the change in speed over time, for some constant acceleration,
#veca = (vecv_f - vecv_i)/(t - 0)#
where we take the initial time to be zero. That is, if you were to plot speed vs. time, the slope is the acceleration if the acceleration is constant.
So, the final speed is:
#color(blue)(vecv_f) = vecat + vecv_i#
#= "0.62 m"/"s"^cancel(2) xx 3.4 cancel"s" + "6.9 m/s"#
#=# #color(blue)("9.0 m/s")#
