Question

Position of a particle moving along x-axis is given by $x=2+8t-4t^2$. The distance travelled by the particle from t = 0 to t = 2 is:

• A. 0
• B. 8
• C. 12
• D. 16

Answer 1

The correct option is B 8

$x=2+8t-4t^2$

$\Rightarrow$ velocity $=\frac{dx}{dt}=0+8-4\left(2\right)t$

or $v=8-8t$

at $t=0v=8m/s$ & $t=1\Rightarrow v=0$

$t=2\Rightarrow v=8-16=-8$

so plotting the graph

we get

there points

$A\equiv (0,8)$

$B\equiv (1,0)$

$C\equiv 2,-8$

distance $=AreaOAB+AreaBOC$

$=\frac{1}{2}OB\times OA+\frac{1}{2}{O}^{\prime }B\times O^{\prime }C=\frac{1}{2}\times 1\times 8+\frac{1}{2}\times (2-1)8=8m$