Question

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

Answer 1

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

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:

$a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t}$ (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)

$V=154.08 \frac{m}{s}$ (4)

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

$V=154.08 \frac{m}{s}=554.68 \frac{km}{h}$ (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)

$V=108 \frac{m}{s}=388.8 \frac{km}{h}$ (6)