Question

The position of a boy moving along a straight line is $x=150\text{m}$ with respect to origin $(x=0)$ at $10.30\text{a.m}$. If he is moving along $+vex-$axis at a constant speed of $6\text{km/hr}$, what will be his position at the time $10.45\text{a.m}$ ?

• A. $750\text{m}$
• B. $1000\text{m}$
• C. $1800\text{m}$
• D. $1650\text{m}$

Answer 1

The correct option is D $1650\text{m}$

Initial position of boy is,
$x_i=150\text{m}$
The time interval for journey of boy is,
$\Delta t=15\text{minute}=\frac{1}{4}\text{hr}$
Boy is moving at a constant speed,
$v=6\text{km/hr}$
Hence displacement of boy along $+vex-$axis is,
$\Rightarrow \text{Displacement}\left(\Delta x\right)=\text{velocity}\times \Delta t$
$\Rightarrow \Delta x=v\Delta t$
$\Rightarrow \Delta x=6\text{km/hr}\times \frac{1}{4}\text{hr}=1.5\text{km}$
$\Rightarrow \Delta x=1.5\times 1000=1500\text{m}$
Hence new position of boy will be,
$x_f=x_i+\Delta x$
$\Rightarrow x_f=150+1500=1650\text{m}$