Physics Problem......?

Question

Starting from rest, a speedboat reaches a speed of 2.8 m/s in 2.7 s. What is the boat's speed after an additional 2.8 s has elapsed, assuming the boat's acceleration remains the same?

Answer 1

a = (Vf - Vo)/T

where

a = acceleration

Vf = final velocity

Vo = initial velocity

T = time

Substituting values,

a = (2.8 - 0)/2.7

a = 1.04 m/sec^2

Use the same formula as above, but this time the following values are true

Vf = final velocity

Vo = 2.8 m/sec.

a = 1.04 m/sec^2

T = 2.8

Substituting appropriate values,

Vf - 2.8 = (1.04)(2.8)

Vf = 2.8 + (1.04)(2.8)

Vf = 5.7 m/sec.

Answer 2

5.7 m/s ?

2.8 * (2.8 + 2.7) / 2.7 = 5.7

Answer 3

First state all of your givens.

Vi=0 m/s (stated by "Starting at rest" in the problem)

Vf=2.8m/s

t1=2.7s

t2=5.5 (2.8+2.7)

a=?

We need to solve for acceleration before finding out the problem.

Use a Kinematics Equation

Vf=Vi+at1

Now Isolate "a"

Vf/a=Vi+t1

a=Vi+t1/Vf

a=t1/Vf

a=2.7/2.8

a=.9642857143 m/s^2

Now solve for t2 using the acceleration you just solved for.

We can use the same kinematics equation. Except t1 is now t2

Vf=Vi+at2

Vf=5.3 mps