A speedboat starts from rest and accelerates at +1.90m/s^2 for 7.00 s. At the end of this time, the boat continues for an additional 6.00 s with an acceleration of +0.529m/s^2. Following this, the boat accelerates at -1.47m/s^2 for 8.00 s.
**What is the total displacement of the boat?
break this into three parts, figure the final velocity, and displacement for each.
A. vf=at=you figure
da=1/2 a t^2 + Vi*t
B. Vi=vf from above.
vfb=vi +a*t
db=viIt + 1/2 a t^2
C do the same.
add the three displacements, da, db, dc
To find the total displacement of the boat, we can use the equations of motion.
First, let's find the displacement during the first 7 seconds when the boat is accelerating at +1.90 m/s^2. We can use the equation:
s = ut + (1/2)at^2
where
s = displacement
u = initial velocity (0 m/s since the boat starts from rest)
t = time (7.00 s)
a = acceleration (+1.90 m/s^2)
Plugging in the values, we get:
s1 = (0)(7.00) + (1/2)(1.90)(7.00)^2
s1 = 0 + 0.5 * 1.90 * (49)
s1 = 0 + 0.5 * 1.90 * 49
s1 = 0 + 0.5 * 94.1
s1 = 0 + 47.05
s1 = 47.05 m
Next, let's find the displacement during the next 6 seconds when the boat is accelerating at +0.529 m/s^2. Using the same equation:
s = ut + (1/2)at^2
where
s = displacement
u = final velocity at the end of the first 7 seconds (which we can find using the equation v = u + at)
t = time for this segment (6.00 s)
a = acceleration (+0.529 m/s^2)
We need to find u, the final velocity at the end of the first segment. Using the equation:
v = u + at
where
v = final velocity at the end of the first segment (which we can find using the equation v = u + at)
u = initial velocity (0 m/s since the boat starts from rest)
a = acceleration (+1.90 m/s^2)
t = time for the first segment (7.00 s)
Plugging in the values, we have:
v = u + at
v = 0 + (1.90)(7.00)
v = 0 + 13.3
v = 13.3 m/s
Now, let's find the displacement during the second segment when the boat is accelerating at +0.529 m/s^2:
s2 = ut + (1/2)at^2
s2 = (13.3)(6.00) + (1/2)(0.529)(6.00)^2
s2 = 79.8 + (0.5)(0.529)(36)
s2 = 79.8 + (0.5)(0.529)(36)
s2 = 79.8 + 0.5 * 0.529 * 36
s2 = 79.8 + 0.5 * 18.984
s2 = 79.8 + 9.492
s2 = 89.292 m
Finally, let's find the displacement during the last segment when the boat is decelerating at -1.47 m/s^2. Again, using the same equation:
s = ut + (1/2)at^2
where
s = displacement
u = final velocity at the end of the second segment (which we can find using the equation v = u + at)
t = time for this segment (8.00 s)
a = acceleration (-1.47 m/s^2)
We need to find u, the final velocity at the end of the second segment. Using the equation:
v = u + at
where
v = final velocity at the end of the second segment (which is the initial velocity for the third segment)
u = final velocity at the end of the second segment (which we found in the previous calculation)
a = acceleration (-1.47 m/s^2)
t = time for the second segment (6.00 s)
Plugging in the values, we have:
v = u + at
v = 13.3 + (-1.47)(8.00)
v = 13.3 + (-11.76)
v = 13.3 - 11.76
v = 1.54 m/s
Now, let's find the displacement during the third segment when the boat is decelerating at -1.47 m/s^2:
s3 = ut + (1/2)at^2
s3 = (1.54)(8.00) + (1/2)(-1.47)(8.00)^2
s3 = 12.32 + (-0.5)(-1.47)(64.00)
s3 = 12.32 + (-0.5)(94.08)
s3 = 12.32 + (-47.04)
s3 = 12.32 - 47.04
s3 = -34.72 m
The total displacement of the boat is the sum of the three displacements:
Total displacement = s1 + s2 + s3
Total displacement = 47.05 + 89.292 + (-34.72)
Total displacement = 101.342
Therefore, the total displacement of the boat is 101.342 meters.