SOLUTION: Show the solutions and answer the given questions. A. A boat is moving smoothly through the waters when the captain noticed a floating log ahead. If the initial velocity of the

Question 1196739: Show the solutions and answer the given questions.
A. A boat is moving smoothly through the waters when the captain
noticed a floating log ahead. If the initial velocity of the boat is 15
m/s and it takes 5.0 s to slow down to 5 m/s, what was the boat’s
average acceleration?
B. What can you say about the acceleration in number 2? What does
it mean?
Found 3 solutions by ikleyn, amoresroy, math_tutor2020:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

(A) an average acceleration is the ratio of the change of speed to the duration of the time interval, by the definition a = = = = -2 m/s^2. The sign " - " (minus) means that the acceleration in this case is, actually, deceleration.

Solved.     ( <<<---=== I just fixed my typo in the post, so now you see the corrected and correct version).

---------------

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Answer by amoresroy(361)   (Show Source): You can put this solution on YOUR website!
The other tutor got the right formula but made a mistake
In substitution.
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the
equation a = Δv/Δt.
Where ∆v = V2 - V1
∆t = t2 - t1
This allows you to measure how fast velocity changes in meters per second squared (m/s^2).
So we have,
a = (5-15)/5-0
a = -2
Answer:
The boat's average acceleration is -2m/s^2.
Negative number means the boat is decelerating
aat 2m/s^2.

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answer: -2 m/s^2
This represents slowing down by 2 m/s per second
I.e. each second the boat's velocity has gone down by 2 m/s

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Work Shown:
vi = initial velocity = 15 m/s
vf = final velocity = 5 m/s
a = acceleration = unknown and what we want to solve for
t = number of seconds that elapsed = 5 seconds

Using one of the four kinematics equations gets us the following
vf = vi + a*t
5 = 15 + a*5
5-15 = 5*a
-10 = 5*a
5*a = -10
a = -10/5
a = -2

If you wanted, you can rewrite that original formula as
a = (vf - vi)/t
which is the same as saying
a = (finalVelocity - initialVelocity)/(endTime - startTime)
effectively we are using something very similar to the slope formula. This is because the acceleration is the rate of change of the velocity.

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