Physics, 29.08.2019 20:30 jamie316

Aspeedboat moving at 31.0 m/s approaches a no-wake buoy marker 100 m ahead. the pilot slows the boat with a constant acceleration of −4.00 m/s2 by reducing the throttle. (a) how long does it take the boat to reach the buoy? s (b) what is the velocity of the boat when it reaches the buoy? m/s

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U = 31 m/s, the initial speed of the boat
s = 100 m, distance to the buoy
a = -4 m/s², the acceleratin (actually deceleration)

Part (a)
Let t = time required to reach the buoy.
Use the formula ut + (1/2)at² = s.
(31 m/s)*(t s) - (1/2)*(4 m/s²)*(t s)² = (100 m)
2t² - 31t + 100 = 0

Solve with the quadratic formula.
t = (1/4) [31 +/- √(31² - 800)]
= 10.92 s or 4.58 s
Before selecting the answer, we should determine the velocity at the buoy.

Part (b)
When t = 10.92 s, the velocity at the buoy is
v = (31 m/s) - (4 m/s²)*(10.92 s) = -12.68 m/s
Because of the negative value, this value of t should be rejected.

When t = 4.58 s, the velocity at the buoy is
v = (31 m/s) - (4 m/s²)*(4.58 s) = 12.68 m/s
This value of t is acceptable.

Answer (to nearest tenth):
(a) The time is 4.6 s
(b) The velocity is 12.7 m/s

Answer from: Quest

c. convergent boundary.

explanation:

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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Aspeedboat moving at 31.0 m/s approaches a no-wake buoy marker 100 m ahead. the pilot slows the boat...

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