1.

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»

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

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}



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