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2.Motion in Straight Line
hard
The position of a body moving along $x$-axis at time $t$ is given by $x=\left(t^2-4 t+6\right) m$. The distance travelled by body in time interval $t=0$ to $t=3 \,s$ is ...... $m$
A$5$
B$7$
C$4$
D$3$
Solution
(a)
$x=t^2-4 t+6$
$\frac{d x}{d t}=2 t-4$
At $t=2 s$, particle is at rest and reverses its position so,
$\left.\begin{array}{l}\left.x\right|_{t=0}=6 \,m \\ \left.x\right|_{t=2 s }=2 \,m \end{array}\right] 4 \,m$
$\left.\begin{array}{l}\left.x\right|_{t=2 s }=2 \,m \\ \left.x\right|_{t=3 s }=3 \,m \end{array}\right] 1 \,m$
Distance $=(4+1) \,m =5 \,m$
Displacement $=4-1=3 \,m$
$x=t^2-4 t+6$
$\frac{d x}{d t}=2 t-4$
At $t=2 s$, particle is at rest and reverses its position so,
$\left.\begin{array}{l}\left.x\right|_{t=0}=6 \,m \\ \left.x\right|_{t=2 s }=2 \,m \end{array}\right] 4 \,m$
$\left.\begin{array}{l}\left.x\right|_{t=2 s }=2 \,m \\ \left.x\right|_{t=3 s }=3 \,m \end{array}\right] 1 \,m$
Distance $=(4+1) \,m =5 \,m$
Displacement $=4-1=3 \,m$
Standard 11
Physics
