A particle moves in a straight line with velocity v ( t ) = | t - 4 | m / s.The distance covered by the particle in 8s is

Question

anonymous · Answer

attachment

abhisar · Answer

Actually this question can be solved in two ways. By integration and by graphical method.

abhisar · Answer

The velocity of the particle is |t-4| this means that the velocity will reach 0 at t=4 s

abhisar · Answer

Then again from t=4s to t=8s velocity will increase from 0 to 4m/s

abhisar · Answer

|dw:1406193224186:dw|

abhisar · Answer

Distance will be equal to the area under the graph i.e `sum of areas of two triangles.`

abhisar · Answer

$\huge\tt \frac{1}{2}\times4\times4$+$\huge\tt \frac{1}{2}\times4\times4$=16m

abhisar · Answer

$\large\sf Integration~Method$

Distance will be equal to $\large\sf\int_{0}^{4}|t-2|+\int_{4}^{8}|t-2|$

abhisar · Answer

$\large \sf \frac{t^2}{2}-4t|^4_0+\frac{t^2}{2}-4t|^8_4~~~~=|-16|=16$

abhisar · Answer

@aaronq can you please check it ?

aaronq · Answer

|dw:1406295523754:dw|

The graph should look like this but the result is correct

abhisar · Answer

oh..Thanx @aaronq !

aaronq · Answer

no probs!