Question

if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?

Answer 1

To calculate the vessel's speed after accelerating for 1 minute, use the formula v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.

Step-by-step explanation:

To determine the final velocity of the vessel after accelerating for 1 minute, we need to know the acceleration and the initial velocity. Since the specific values are not provided in your question, let's consider the example mentioned regarding the freight train that accelerates at a rate of 0.0500 m/s² for 8.00 minutes, starting with an initial velocity of 4.00 m/s. This information allows us to apply the formula for final velocity:

v = v0 + at

Using this formula, we can solve for the vessel's speed after acceleration. However, without the exact values of acceleration and initial velocity for the vessel, we cannot calculate a numerical answer. If, for example, a vessel accelerates at the same rate as the freight train for 1 minute (which is 0.0500 m/s²), and begins at the same initial velocity of 4.00 m/s, the formula would be:

v = 4.00 m/s + (0.0500 m/s²) × (60 s)

v = 4.00 m/s + 3 m/s = 7.00 m/s

This means the vessel would have a final velocity of 7.00 m/s after accelerating for 1 minute.

Answer 2

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Step-by-step explanation:

The complete question is written below:

In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of
`1.80 m/s^(2)`, it would go from rest to its top speed in 85.6 s.

a) What was the speed of the vessel?

b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute?

Calculate the answers in both meters per second and kilometers per hour

a) The average acceleration
`a_(av)` is expressed as:

`a_(av)=(\Delta V)/(\Delta t)=(V-V_(o))/(\Delta t)` (1)

Where:

`a_(av)=1.80 m/s^(2)`

`\Delta V` is the variation of velocity in a given time
`\Delta t`, which is the difference between the final velocity
`V` and the initial velocity
`V_(o)=0` (because it starts from rest).

`\Delta t=85.6 s`

Isolating
`V` from (1):

`V=a_(av)\Delta t + V_(o)` (2)

`V=(1.80 m/s^(2))(85.6 s) + 0 m/s` (3)

`V=154.08 (m)/(s)` (4)

If
`1 km=1000m` and
`1 h=3600 s` then:

`V=154.08 (m)/(s)=554.68 (km)/(h)` (4)

b) Now we need to find the final velocity when
`\Delta t=1 min=60 s`:

`V=(1.80 m/s^(2))(60 s) + 0 m/s` (5)

`V=108 (m)/(s)=388.8 (km)/(h)` (6)