- Home
- Standard 9
- Science
Joseph jogs from one end $A$ to the other end $B$ of a straight $300\, m$ road in $2\, min$ $30\, sec$ and then turns around and jogs $100 \,m$ back to point $C$ in another $1\, min$. What are Joseph's average speeds and velocities in jogging from $A$ to $C$ ?
$904 \,m s ^{-1}$ and $952\, m s ^{-1}$
$0.904 \,m s ^{-1}$ and $1.952\, m s ^{-1}$
$1.904 \,m s ^{-1}$ and $0.952\, m s ^{-1}$
$0.809 \,m s ^{-1}$ and $0.599\, m s ^{-1}$
Solution
Total Distance covered from $AC = AB + BC$
$=300+200 \,m$
Total time taken from $A$ to $C =$ Time taken for $AB$ $+$ Time taken for $BC$
$=(2 \times 60+30)+60\, s$
$=210 \,s$
Therefore, Average Speed from $AC =$ Total Distance $/$ Total Time
$=400 / 210\,m s ^{-1}$
$=1.904 \,m s ^{-1}$
Displacement $(s)$ from $A$ to $C = AB – BC$
$=300-100 \,m$
$=200 \,m$
Time $(t)$ taken for displacement from $AC =210 \,s$
Therefore, Velocity from $A C=$ Displacement $(s) /$ Time $(t)$
$=200 / 210 \,m s ^{-1}$
$=0.952\, m s ^{-1}$
