Thank you for reporting, we will resolve it shortly
Q.
The variation of velocity of a particle moving along a
straight line is shown in figure. The distance traversed by the particle in 4 second is
Motion in a Straight Line
Solution:
Distance travelled in $4 s =$ Area under the curve
$=$ area of $\Delta A O H+$ area of rectangle $A B G H+$ area of $\Delta B C C'$
$+$ area of rectangle $C' D E G$
$=\frac{20 \times 1}{2}+20 \times 1+\frac{10 \times 1}{2}+10 \times 2=55 \,m$
