Question

a motorboat starting from rest on a lake accelerate in straight line at the constant rate of 3.0 M/s2 for 6 sec how far does the boat travel during this time class 9

Answer 1

Answer:

54 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 0 m/s [As it starts from rest]
• Acceleration (a) = 3 m/s²
• Time taken (t) = 6 s

We've been asked to calculate the distance travelled by the boat.

Here, we have u, a and t. So, we can apply here the 2nd equation of motion. That is,

$\dashrightarrow \quad \boxed{\rm {s =ut + \dfrac{1}{2}at^2}}$

• s denotes distance
• u denotes initial velocity
• t denotes time
• a denotes acceleration

Now, substitute the values ;

$\dashrightarrow \quad \rm { s= 0(6) + \dfrac{1}{2}(3)(6)^2 \; m}$

$\dashrightarrow \quad \rm { s= 0(6) + \dfrac{1}{2}(3)(36) \; m}$

$\dashrightarrow \quad \rm { s= (3)(18) \; m}$

$\dashrightarrow \quad\underline{\boxed {\bf {s =54 \; m}}}$

∴ The boat travels 54 m during this time.

$\rule{200}2$

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Equations of motion :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes distance

Answer 2

$\huge\purple{ \underline{ \boxed{\mathbb\colorbox{cyan}{\red{QUESTION : }}}}}$

A motorboat starting from rest on a lake accelerate in straight line at the constant rate of 3.0 $\sf ms^{-2}$ for 6 seconds .How far does the boat travel during this time .

$\huge\purple{ \underline{ \boxed{\mathbb\colorbox{cyan}{\red{ANSWER : }}}}}$

$\bf Given\begin{cases}\bf Initial\:Velocity\:(u)=0\\\\\bf Acceleration\:(a)=3\: ms^{-2} \\\\\bf Time\: Taken\:(t)=6\:s\end{cases}$

$\green{ \large \underline{ \mathbb{\underline{FORMULA \: USED: }}}}$

$\boxed{\displaystyle\bf s=ut+\frac{1}{2}at^{2} }$

$\bf where\begin{cases}\bf s\:\;denotes\: distance\\\\\bf u\:\;denotes\: initial\: velocity\: \\\\\bf a\:\;denotes\: acceleration\\\\\bf t\:\;denotes\: time\: taken\end{cases}$

$\longrightarrow\displaystyle \bf s=0(6)+\frac{1}{2} (3)(6)^{2} \:\:\:\\\\\longrightarrow\displaystyle \bf s=0+\frac{1}{2} \;\times\;3\;\times\;36\\\\\longrightarrow\displaystyle \bf s=\frac{108}{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\\\\\longrightarrow\displaystyle \underline{\boxed{\bf\:s=54 \;m}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$

Boat travels 54 m during this time .

$\Large{\mathbb{LEARN\;MORE :}}$

Equations of motion :

$\bf v\:=\:u\:+\:at\:\:\:\:\:\:\:\:\\\\\bf v^{2} \bf-u^{2} \:=\:2as\:\:\:\:\:\\\\\displaystyle\bf s\:=\:ut \:+\:\frac{1}{2} at^{2}$