7. MOTION
medium

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 $C$ ?

A

$904 \,m s ^{-1}$ and $952\, m s ^{-1}$

B

$0.904 \,m s ^{-1}$ and $1.952\, m s ^{-1}$

C

$1.904 \,m s ^{-1}$ and $0.952\, m s ^{-1}$

D

$0.809 \,m s ^{-1}$ 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 $(t)$

$=200 / 210 \,m s ^{-1}$

$=0.952\, m s ^{-1}$

Standard 9
Science

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