velocity, PLEASE help!!!!!!!!!!!!!!!!!?

Question

Please help!

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!!!!!!!!!!!!!!!!!!

i posted this earlier but all the answers i got were wrong. i checked them and the computer tells if your right or wrong but not how to do them or what the right answer is so...
please, PLEASE, help me. i desperately need it.

i sincerly appreciate it!

Answer 1

The answers are
A.)3.06 m/s
B.) 212.92 meters

But to survive physics you need to understand how i got that answer.

To solve this you need to know two physics equations:

1.) Xf = Xi + Vi*t + .5a(t^2)
Xf and Xi are final and initial displacement, Vi is initial velocity, t is time, a is acceleration. t^2 means time squared.

2.) Vf = Vi + at

You need to split the problem into three for each part.

Part 1: Velocity at 21.15 seconds
Dont let it trick you, 21.15 seconds is the total time (6.55+5.8+8.8)

We use Vf = Vi + at and solve for each velocity.
The boat starts at rest (Vi=0 m/s), acceleration = 2.01, time =6.55.
So we punch everything in (Vf=0+ 2.01*6.55) and get 13.17 m/s
Now the boat changes its acceleration to .518 now it has an initial velocity of 13.17
So again, (Vf=13.17 + .518*5.8) and we get 16.17 m/s
the boat changes its acceleration again to -1.49 but its initial velocity is now 16.17m/s
(Vf = 16.17 + (-1.49*8.8) ) we get 3.06 m/s as our final velocity.

Part 2: Final Displacement.
Again we divide the problem into three.
We use Xf=Xi + Vi*t + .5a(t^2) for this part.
The boat starts from 0 displacement.
The first stretch:
(Xf = 0 + 0*6.55 + .5(2.01)(6.55^2) ) we get 43.12 m.
Second Stretch,
the initial displacement is now 43.12 m, the initial velocity on the second stretch was 13.17 m/s, and the acceleration is now .518 m/s squared. It does this for 5.8 seconds
(Xf = 43.12 + 13.17*5.8 + .5(.518)(5.8^2) )
We get 128.22 m
Third Stretch:
the initial displacement is now 128.22 m, the initial velocity on the third stretch was 16.17 m/s, and the acceleration is now -1.49 m/s squared. It does this for 8.8 seconds
(Xf = 128.22 + 16.17*8.8 + .5(-1.49)(8.8^2) )
We get a final displacement of 212.92 meters

Answer 2

v = u + at; where u = initial velocity, a = acceleration, and t = time of acceleration. v = terminal velocity at the end of t while accelerating at a. Use this SUVAT equation interatively

For example, when a1 = 2.01 m/sec^2 t1 = 6.55 sec and u1 = 0 the rest velocity. Plug these numbers into v1 = u1 + a1t1 and solve for v1, the terminal velocity after 6.55 sec.

Now we have v2 = v1 + a2t2; where v1 is the initial velocity you just solved starting the second leg with a2 = .518 m/sec^2 for t2 = 5.8 sec. Solve for v2.

Finally, v3 = v2 + a3t3; where v2 you solved for earlier, a3 = - 1.49 m/sec^2 and t3 = 8.8 sec Solve for v3, which is your answer for a).

Or you can be really clever in that v3 = v2 + a3t3 = (v1 + a2t2) + a3t3 = (u + a1ta) + a2t2 + a3t3 = a1t1 + a2t2 + a3t3. In other words, the final velocity after t = t1 + t2 + t3 = 21.15 sec is just the sum of the accelerations times each time of acceleration.

For b) note that s1 = s0 + 1/2 a1t1^2; where s0 = 0 the initial distance traveled and s1 is the distance traveled during the first leg. Thus s2 = s1 + 1/2 a2t2^2 and s3 = s2 + 1/2 a3t3^2.

Following the same line of reasoning for a), we can find that s3 = 1/2 a1t1^2 + 1/2 a2t2^2 + 1/2 a3t3^2 = 1/2 (a1t1^2 + a2t2^2 + a3t3^2) = the total displacement of the boat. Here the first term is the distance of the first leg, to which the distance of the second leg is added, and the third leg is added to the first two legs.

Answer 3

(a) t=21.25s is the end of the journey so we have to follow through the whole trip.
first phase:
v =u +at
v = 0 + (2.01 x 6.55)
v = 13.1655 m/s

second phase:
v =u + at
v = 13.1655 + (0.518 x 5.8)
v = 16.1699 m/s

third phase(decelleration):
v = u - at
v = 16.1699 -(1.49 x 8.80)
v = 3.0579 m/s

(b)again we have to find the displacement in each phase and add them all up:
first phase:
s = ut + 1/2at^2
s1 = 0 + (1/2 x 2.01 x 6.55^2)
s1 = 43.117 m

second phase:
s2 = (13.1655 x 5.8) + (1/2 x 0.518 x 5.8^2)
s2 = 76.3599 + 8.7128
s2 = 85.0727 m

third phase:
s3 = (16.1699 x 8.8) - (1/2 x 1.49 x 8.8^2)
s3 = 142.295 - 57.6928
s3 = 84.6023m

total displacement = s1 + s2 + s3
= 212.792 m