What is the displacement for a particle that starts at #"3.89 m/s"#, and has a constant acceleration of #"3.55 m/s"^2# for #"4.5 s"#?

#Deltavecx = "46.7 m"#

You should familiarize yourself with the 1D kinematics equations and when to use them:
https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas

Given a constant acceleration and a (clearly constant) initial speed, we can first define the acceleration as the slope of a speed vs. time graph:

#veca = (vecv_f - vecv_i)/t#

where we define initial time as zero. #vecv# is the velocity but we look at only one direction (forward).

Rearrange to get the final speed (i.e. the speed at any time #t#):

#vecv_f = vecat + vecv_i#

If we start from here we can derive an equation for the change in position as a function of acceleration, initial velocity, and time.

From Calculus, velocity is the change in position over time, but the initial speed never changes. So, with #vecx# being the current position:

#(dvecx)/(dt) = vecat + vecv_i#

Multiply over the #dt#:

#dvecx = (vecat + vecv_i)dt#

As long as acceleration remains constant, we can integrate both sides to get:

#int_(vecx_1)^(vecx_2) dvecx = int_(0)^(t) (vecat + vecv_i)dt#

#=> Deltavecx = vecaint_(0)^(t) tdt + vecv_iint_(0)^(t) dt#

Using the power rule from Calculus, the change in position then becomes:

#barul(|stackrel(" ")(" "Deltavecx = 1/2 vecat^2 + vecv_it" ")|)#

Of course, if you aren't in a Calculus-based course, you can simply use this equation as it is, as long as you know what it means physically... And you should verify that this equation is correct:
https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas

So, the displacement (i.e. the change in position) is:

#color(blue)(Deltavecx) = 1/2 ("3.55 m/s"^2)("4.15 s")^2 + ("3.89 m/s")("4.15 s")#

#=# #color(blue)("46.7 m")#