Question

1 poir Q5. An object travels 5 km in 1h and then 20 km in 4 h. What is the average speed of the object? *​

Answer 1

Provided that:

• An object travels 5 km in 1 hr.

• Then the object travels 20 km in 4 hr.

To calculate:

• The average speed

Solution:

• The average speed = 5 mps

Using concept:

• Average speed formula

Using formula:

• ${\small{\underline{\boxed{\pmb{\sf{v \: = \dfrac{s + s'}{t + t'}}}}}}}$

Where, v denotes average speed, s denotes distance one, s′ denotes second distance, t denotes time one and t′ denotes second time.

Required solution:

$:\mapsto \sf v \: = \dfrac{s + s'}{t + t'} \\ \\ :\mapsto \sf v \: = \dfrac{5+20}{1+4} \\ \\ :\mapsto \sf v \: = \dfrac{25}{5} \\ \\ :\mapsto \sf v \: = 5 \: ms^{-1} \\ \\ :\mapsto \sf Average \: speed \: = 5 \: ms^{-1}$

Know more:

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