A motor boat is traveling at a speed of 2.4 m/s shuts off its engine at t=0. How far does it travel before coming to a rest if it is noted that after 3 seconds its speed has dropped to half of its original value? Assume that the drag force of the water is proportional to v. (R=-bv)
To determine the distance traveled by the motor boat before coming to a rest, we can use the concept of acceleration and the equation of motion.
1. First, let's define the variables given in the problem:
- Initial speed (u) = 2.4 m/s
- Time when the engine shuts off (t1) = 0 s
- Time when the speed drops to half (t2) = 3 s
- Final speed (v) = u/2 (because the speed has dropped to half)
2. We need to find the acceleration (a) experienced by the boat after the engine shuts off. By using Newton's second law of motion (F = ma) with the drag force equation (F = -bv), we can equate them:
- ma = -bv
- a = (-bv)/m
3. The equation of motion relates displacement (s), initial speed (u), final speed (v), acceleration (a), and time (t):
- s = ut + (1/2)at^2
4. Now, we can find the acceleration (a) at time t = t2:
- a = (-bv)/m
- (-bv)/m = (v - u)/t2 [Substituting a = (v - u)/t2]
- (-b(2.4))/m = ((2.4/2) - 2.4)/3 [Substituting v = u/2 and t2 = 3]
- (-2.4b)/m = (-1.2 - 2.4)/3
- (-2.4b)/m = -3.6/3
- (-2.4b)/m = -1.2
- b/m = 1.2/2.4
- b/m = 0.5
5. We have determined the value of b/m, which is 0.5, representing the drag force coefficient.
6. Now, we can find the acceleration (a) at time t = 0:
- a = (-bv)/m
- (-bv)/m = (v - u)/t1 [Substituting a = (v - u)/t1]
- (-b(2.4))/m = (0 - 2.4)/0
- (-2.4b)/m = -2.4/0
- The above equation is indeterminate, so we need to use the limit concept:
- lim((-2.4b)/m) as t->0 = 2.4
7. Now that we know the acceleration (a) is 2.4 m/s², we can substitute the values into the equation of motion:
- s = ut + (1/2)at^2
- s = 2.4(0) + (1/2)(2.4)(0)^2 [Substituting u = 2.4, t = 0, and a = 2.4]
- s = 0
Therefore, the motor boat does not travel any distance before coming to rest since it shuts off its engine at t = 0 seconds.