A speedboat starts from rest and accelerates at +2.07 m/s2 for 7.00 s. At the end of this time, the boat continues for an additional 6.00 s with an acceleration of +0.523 m/s2. Following this, the boat accelerates at -1.55 m/s2 for 8.00 s. What is the velocity of the boat at t = 21.0 s?
The velocity at the end of the first interval is (2.07 m/s^2)*(7.00s) = 14.49 m/s
The velocity increases by an amount (0.523 m/s^2)(6.00s) = 3.14 m/s
during the second interval. This brings the velocity up to 17.63 m/s
During the third interval, the velocity decreases by (1.55 m/s^2)*(8.00 s) = 12.40 m/s
Now complete the final step. Subtract 12.40 from 17.63 m/s
To find the velocity of the boat at t = 21.0 s, we need to calculate the total displacement of the boat during the given time intervals, and then use the equation of motion to determine the final velocity.
1. For the initial 7.00 s when the boat accelerates at +2.07 m/s^2:
Using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can find the final velocity at the end of this interval.
u = 0 m/s (since the boat starts from rest)
a = +2.07 m/s^2
t = 7.00 s
v = 0 + (2.07)(7.00) = 14.49 m/s
2. For the next 6.00 s when the boat accelerates at +0.523 m/s^2:
Using the same equation of motion, we can calculate the final velocity at the end of this interval.
u = 14.49 m/s (since this is the velocity from the first interval)
a = +0.523 m/s^2
t = 6.00 s
v = 14.49 + (0.523)(6.00) = 17.543 m/s
3. Lastly, for the final 8.00 s when the boat accelerates at -1.55 m/s^2:
Using the equation of motion once again, we can find the final velocity at the end of this interval.
u = 17.543 m/s (since this is the velocity from the second interval)
a = -1.55 m/s^2
t = 8.00 s
v = 17.543 + (-1.55)(8.00) = 5.343 m/s
So, the velocity of the boat at t = 21.0 s is 5.343 m/s.