Question

A motor boat starrting from rest on a lake aocelration in a straight line at a constant rate of 3.0 meter/square second for 8.0s .how far dose the boat travel during this tim?​

Answer 1

GIVEN:-

• $\rm{Initial\:Velocity(u)=0m/s}$

• $\rm{Acceleration=3m/s^{-2}}$

• $\rm{Time=8.0s}$

TO FIND:-

• Distance travelled by boat during this time.

FORMULAE USED:-

• ${\boxed{\rm{\blue{S=ut+\dfrac{1}{2}\times{a}\times{(t)^2}}}}}$

Where,

S= Distance

U= initial Velocity

t= Time

a= Acceleration

Now,

$\implies\tt{S=ut+\dfrac{1}{2}\times{a}\times{(t)^2}}$

$\implies\tt{S=0\times{8}+\dfrac{1}{2}\times{3}\times{(8)^2}}$

$\implies\tt{S=0+\dfrac{1}{\cancel{2}}\times{3}\times{{\cancel{64}}}}$

$\implies\tt{S=0+96}$

$\implies\tt{S=96m}$

Hence, The distance travelled by motor boat is 96m.

EXTRA INFORMATION

• When object starts from the rest than we take Initial Velocity as zero.

• Velocity is a vector quantity because it has a magnitude as well as Direction.

• Acceleration is also a Vector quantity.

Answer 2

Hi there!

Your Answer :- 96m.

$\boxed{Given\ratio - }$

Initial Velocity = 0m/s.

Time = 8s.

Acceleration = 3m/s-²

$\boxed{To \: find \ratio - }$

Distance traveled.

$\boxed{Solution\ratio - }$

Apply 2nd Equation of Motion :-

$\rm\red{s = ut + \dfrac{1}{2} a {t}^{2} }$

Put the values.

$\rm\green{s = 0 \times 8 + \dfrac{1 \times 3 \times {8}^{2} }{2} }$

$\rm\blue{s = 0 + \dfrac{1}{2} \times 3 \times 64}$

$\rm\purple{s = 0 + 96}$

$\sf\fbox\pink{s = 96 \: m}$

So, the answer is 96 m.