A boat moves slowly out of a marina (so as to not leave a wake) with a speed of 1.50 m/s. As soon as it passes the breakwater, leaving the marina, it throttles up and accelerates at 2.40 m/s/s.
a) How fast is the boat moving after accelerating for 5 seconds?
b) How far did the boat travel during that time?
The correct answer and explanation is :
We will use kinematic equations to solve this problem.
Given Data:
- Initial velocity, ( v_i = 1.50 ) m/s
- Acceleration, ( a = 2.40 ) m/s²
- Time, ( t = 5.00 ) s
(a) Final Velocity after 5 seconds
Using the kinematic equation:
[
v_f = v_i + at
]
[
v_f = 1.50 + (2.40 \times 5.00)
]
[
v_f = 1.50 + 12.00 = 13.50 \text{ m/s}
]
So, the boat’s velocity after 5 seconds is 13.50 m/s.
(b) Distance Traveled
Using the kinematic equation:
[
s = v_i t + \frac{1}{2} a t^2
]
[
s = (1.50 \times 5.00) + \frac{1}{2} (2.40) (5.00)^2
]
[
s = 7.50 + \frac{1}{2} (2.40) (25)
]
[
s = 7.50 + (1.20 \times 25)
]
[
s = 7.50 + 30.00 = 37.50 \text{ m}
]
So, the boat travels 37.50 meters during this time.
Explanation:
The boat initially moves at a slow speed of 1.50 m/s to avoid making a wake in the marina. As soon as it exits the breakwater, it starts accelerating at 2.40 m/s². To determine how fast it is moving after 5 seconds, we use the equation ( v_f = v_i + at ), which accounts for the constant acceleration over time. This gives us a final velocity of 13.50 m/s.
To find the distance traveled, we use the equation ( s = v_i t + \frac{1}{2} a t^2 ), which considers both the initial motion and the effect of acceleration. The boat covers 37.50 meters in 5 seconds. These calculations show how acceleration significantly increases both speed and distance traveled over time.