Question

A particle moves on a straight line and its velocity is given as $v=(3-t)m/s$ Find distance travelled by it is first 5 seconds.

• A. $2.5m$
• B. $4.5m$
• C. $6.5m$
• D. $8m$

Answer 1

The correct option is C $6.5m$

The relation is $\frac{dx}{dt}=v$ where $x$ is displacement and $v$ is velocity which is given as $v=3-t$

so we have $dx=vdt$

after $3sec$ the direction of motion will get reversed.

so

on integrating both sides we get ${\int }_{X=0}^{X}dx={\int }_{t=0}^{t=3}(3-t)dt=9-\frac{9}{2}=4.5meter$

now integrating both sides for $t=3$ to $t=5$

we have the $displacement$ as

${\int }_{X=4.5}^{X}dx={\int }_{t=3}^{t=5}(3-t)dt=-2meter$

so the net distance will be $X=\left(2\right)+4.5=6.5meter$