Question

Salma takes 20 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 5 m/s, calculate the distance between her house and the school. (this is a class 7th question)

Answer 1

Answer:

Provided that:

• Time taken = 20 minutes
• Speed = 5 m/s

To calculate:

• The distance

Solution:

• The distance = 6000 m

Knowledge required:

• SI unit of distance = m
• SI unit of time = sec
• SI unit of speed = m/s

Using concepts:

• Formula to convert min-sec
• Formula to find distance

Using formulas:

• 1 min = 60 sec
• Distance = Speed × Time

Required solution:

~ Firstly let us convert min into sec!

→ 1 min = 60 sec

→ 20 min = 20 × 60 sec

→ 20 min = 1200 seconds

• Henceforth, converted!

~ Now let's calculate distance!

→ Distance = Speed × Time

→ Distance = 5 × 1200

→ Distance = 6000 m

• Henceforth, solved! $\:$

Difference between distance and displacement:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Distance&\bf Displacement\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf Path \: of \: length \: from \: which &\sf The \: shortest \: distance \: between \\ \sf \: object \: is \: travelling \: called \: distance. &\sf \: the \: initial \: point \: \& \: final \\ &\sf point \: is \: called \: displacement. \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \end{array}}\end{gathered}$

Difference between speed and velocity:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Speed&\bf Velocity\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf The \: distance \: travelled \: by &\sf The \: distance \: travelled \: by \\ \sf \: a \: body \: per \: unit \: time&\sf \: a \: body \: per \: unit \: time \\ &\sf in \: a \: given \: direction \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \\\\\sf Speed \: = \dfrac{Distance}{Time} &\sf Velocity \: = \dfrac{Displacement}{Time} \end{array}}\end{gathered}$

Answer 2

Answer:

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}\end{gathered}$

• ➳ Time taken by Salma from her house to reach her school on a bicycle = 20 minutes
• ➳ Speed of Bicycle = 5 m/s

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{To Find :}}}}}}\end{gathered}$

• ➳ The distance between house and the school.

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Using Formula :}}}}}}\end{gathered}$

$\quad\dag\underline{\boxed{\sf{\purple{Distance = Speed \times time}}}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}\end{gathered}$

$\bigstar{\underline{\underline{\pmb{\frak{\red{Here : }}}}}}$

$\quad\dashrightarrow\sf{Speed = 5 \: m/s}$

$\quad \dashrightarrow\sf{Time = 20 \: minutes }$

$\rule{200}2$

$\bigstar{\underline{\underline{\pmb{\frak{\red{Firstly, \: converting \: the \: time \: into \: second\: : }}}}}}$

$\quad{: \implies{\sf{1 \: minute = 60 \: seconds}}}$

• 20 minutes into seconds

$\quad{: \implies{\sf{20 \: minute = 60 \: \times 20}}}$

$\quad{: \implies{\sf{20 \: minute = 1200 \: seconds}}}$

$\quad\dag{\underline{\boxed{\sf{Time = 1200 \: seconds }}}}$

• Hence, The time is 1200 seconds..

$\rule{200}2$

$\bigstar{\underline{\underline{\pmb{\frak{\red{Now,Finding \: the \: distance : }}}}}}$

$\quad{ : \longmapsto{\sf{Distance = Speed \times time}}}$

• Substituting the values

$\quad{ : \longmapsto{\sf{Distance =1200\times 5}}}$

$\quad{ : \longmapsto{\sf{Distance =6000 \: m}}}$

$\quad\dag{\underline{\boxed{\sf{Distance = 6000 \: m }}}}$

• Henceforth,The Distance between Salma's house to the school is 6000 m.

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Learn More :}}}}}}\end{gathered}$

$\bigstar{\underline{\underline{\pmb{\frak{Formulas : }}}}}$

$\quad\dag\underline{\boxed{\sf{\red{Distance = Speed \times time}}}}$

$\quad\dag{\underline{\boxed{\sf{\red{Time = \dfrac{Distance}{Speed}}}}}}$

$\quad\dag{\underline{\boxed{\sf{\red{Speed = \dfrac{Distance}{Time}}}}}}$

$\rule{200}2$

$\bigstar{\underline{\underline{\pmb{\frak{Units : }}}}}$

$\quad\dashrightarrow{\bf{\pink{Time}} : \sf \purple{ Seconds, minutes, hours}}$

$\quad\dashrightarrow \bf{\pink{Distance}}: {\sf{\purple{meter, kilometer}}}$

$\quad\dashrightarrow \bf{\pink{Speed}} : \sf{\purple{ km/ hr, m /sec}}$

$\rule{200}2$

$\bigstar{\underline{\underline{\pmb{\frak{Conversion \: of \: Units: }}}}}$

$\quad\leadsto\sf{\green{1 \: km/hr} = \blue{5/18 \: metre/second}}$

$\quad \leadsto\sf \green{1 \: metre/second }= \blue{18/5 \: km/hr}$

$\quad\leadsto\sf \green{1 \: Km/hr}= \blue{ 5/8 \: mile/hr}$

$\quad \leadsto\sf \green{1 \: mile/hr} = \blue{22/15 \: foot/second }$

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