A speedboat starts from rest and accelerates at +1.98 m/s2 for 8.25 s. At the end of this time, the boat continues for an additional 5.58 s with an acceleration of +0.348 m/s2. Following this, the boat accelerates at -1.42 m/s2 for 7.22 s. (a) What is the velocity of the boat at t = 21.05 s? (b) Find the total displacement of the boat.
To solve this problem, we can use the equations of motion. Let's break down the problem into different intervals and calculate the velocity and displacement at each interval.
Interval 1: From t=0 to t=8.25 s with an acceleration of +1.98 m/s^2
Using the equation of motion: v = u + at
We know the initial velocity u = 0 (starting from rest), acceleration a = +1.98 m/s^2, and time t = 8.25 s.
Substituting the values, we get:
v1 = 0 + (1.98)(8.25)
v1 = 16.335 m/s
Interval 2: From t=8.25 s to t=13.83 s with an acceleration of +0.348 m/s^2
Using the same equation of motion:
v2 = v1 + a*t2
Where v1 is the velocity at the end of the first interval, a is the acceleration, and t2 is the time for the second interval.
Substituting the values:
v2 = 16.335 + (0.348)(5.58)
v2 = 18.795 m/s
Interval 3: From t=13.83 s to t=21.05 s with an acceleration of -1.42 m/s^2
Using the same equation of motion:
v3 = v2 + a*t3
Where v2 is the velocity at the end of the second interval, a is the acceleration, and t3 is the time for the third interval.
Substituting the values:
v3 = 18.795 + (-1.42)(7.22)
v3 = 7.3414 m/s
(a) Velocity at t=21.05 s:
The velocity at this time is given by v3, which we just calculated:
v = v3
v = 7.3414 m/s
(b) Total displacement:
The total displacement is the sum of the individual displacements during each interval. We can calculate it using the equation:
s = ut + (1/2)at^2
For each interval, we need to calculate the displacement separately, and then add them up.
Displacement during interval 1:
s1 = (0)(8.25) + (1/2)(1.98)(8.25)^2
s1 = 68.07675 m
Displacement during interval 2:
s2 = (16.335)(5.58) + (1/2)(0.348)(5.58)^2
s2 = 136.50384 m
Displacement during interval 3:
s3 = (18.795)(7.22) + (1/2)(-1.42)(7.22)^2
s3 = 76.837276 m
Total displacement:
s_total = s1 + s2 + s3
s_total = 68.07675 + 136.50384 + 76.837276
s_total = 281.417066 m
Therefore, the velocity at t=21.05 s is 7.3414 m/s and the total displacement of the boat is 281.417066 m.
To find the velocity of the boat at t = 21.05 s, you need to find the total acceleration and then use the kinematic equation for velocity.
Step 1: Find the total acceleration.
The boat has three different accelerations in different time intervals. To find the total acceleration, you need to find the sum of the accelerations for each time interval.
Acceleration in the first time interval: +1.98 m/s^2
Acceleration in the second time interval: +0.348 m/s^2
Acceleration in the third time interval: -1.42 m/s^2
Total acceleration = 1st interval's accel. + 2nd interval's accel. + 3rd interval's accel.
Total acceleration = 1.98 m/s^2 + 0.348 m/s^2 + (-1.42 m/s^2)
Total acceleration = 0.908 m/s^2
Step 2: Use the kinematic equation for velocity.
The kinematic equation for velocity is:
v = u + at
where:
v = final velocity
u = initial velocity (which is 0 in this case)
a = acceleration
t = time
At t = 21.05 s, you want to find the velocity, so:
v = 0 + (0.908 m/s^2)(21.05 s)
v = 0.908 m/s^2 * 21.05 s
v ≈ 19.08 m/s
So, the velocity of the boat at t = 21.05 s is approximately 19.08 m/s.
Now, to find the total displacement of the boat, you need to find the displacement for each time interval and then sum them up.
Displacement in the first time interval:
Using the equation:
s = ut + (1/2)at^2
where:
s = displacement
u = initial velocity (which is 0 in this case)
a = acceleration
t = time
s1 = 0 + (0.5)(1.98 m/s^2)(8.25 s)^2
s1 = 0.5 * 1.98 m/s^2 * (8.25 s)^2
s1 ≈ 67.16 m
Displacement in the second time interval:
Using the same equation for displacement:
s2 = 0 + (0.5)(0.348 m/s^2)(5.58 s)^2
s2 ≈ 2.41 m
Displacement in the third time interval:
s3 = 0.5 * (-1.42 m/s^2) * (7.22 s)^2
s3 ≈ -36.42 m
Total displacement = displacement in 1st interval + displacement in 2nd interval + displacement in 3rd interval
Total displacement = 67.16 m + 2.41 m + (-36.42 m)
Total displacement ≈ 33.15 m
So, the total displacement of the boat is approximately 33.15 m.
Vf = Vo + at,
Vf = 0 + 1.98m/s^2 * 8.25s = 16.34 m/s.
d1 = Vo*t + 0.5a*t^2,
d1 = 0 + 0.5*1.98*(8.25)^2 = 67.38 m.
Vf = Vo + at,
Vf=16.34 + 0.348 * 5.58s.=18.28 m/s^2.
d2 = Vo*t + 0.5a*t^2,
d2=16.34*5.58 + 0.5*0.348*(5.58)^2 = 96.6 m.
Vf = Vo + at,
Vf = 18.28 - 1.42*7.22 = 8.03 m/s.
d3=18.28 - 0.5*1.42*(7.22)^2=-18.73 m.
a. 8.03 m/s.
b. D = 67.4 + 96.6 - 18.73 = 145.3 m.