please help!?

Question

A speedboat begins at the rest and accelerates at +2.01 m/s2 for 6.55 s. At the end of this time, the boat continues for an additional 5.80 s with an acceleration of +0.518 m/s2. Following this, the boat accelerates at -1.49 m/s2 for 8.80 s.
(a) What is the velocity of the boat at t = 21.15 s?

(b) Find the total displacement of the boat.

Thank you so much!!!!!!!!!!!!!!!!!!

as long as you kinda show some work, you don't need to explain.

Answer 1

(a) V=V0+at
This is an iterative problem
V1=V0+a1t1 = 0 + 2.01 x 6.55=13.2m/s
V2=V1+a2t2 = 13.2 + 0.518 x 5.80 = 16.17m/s
V3=V2+a3t3= 16.17 - 1.49 x 8.80=3.06 m/s

Caution, here is a trick -
t= 6.55 + 5.88 + 8.80=21.23s
dt= 21.23-21.15=0.08s
t3'=t3 - 0.08=8.72 s

And finally the speed at t=21.15 s is
V3'=V2+a3t3'= 16.17 - 1.49 x 8.72=3.18 m/s


(b)
In this case
S=S1+S2+....+Sn
and since
S= S0 + V0t + 0.5 at^2 we have

S1= S0 + V0t1 + 0.5 a1 t1^2=
S1= 0 + 0 + 0.5 x 2.01 (6.55)^2=
S1=43.1m

S2= S1+ V1 t2 + 0.5 a2 t2^2
S2= 43.1 + 13.2 x 5..8 + 0.5 x 0.518 (5.8)^2=
S2= 128.4 m

finally
S3= S2+ V2 t3 - 0.5 a3( t3)^2

Answer 2

speed = (+2.01 m/s2)*(6.55 s) + (+0.518 m/s2)*(5.80 s) + (-1.49 m/s2)*(8.80 s) = 3,058 m/s
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displacement = 0.5*(+2.01 m/s2)*(6.55 s)^2 + (+2.01 m/s2)*(6.55 s)*(5.80 s) + 0.5*(+0.518 m/s2)*(5.80 s)^2 + [(+2.01 m/s2)*(6.55 s) + (+0.518 m/s2) * (5.80 s)]*(8.80 s) + 0.5*(-1.49 m/s2)* (8.80 s)^2 = 212,79 m
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