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The displacement of a particle is given by $x = (t - 2)^{2}$ where x is in metre and t in second. The distance covered by the particle in first 4 seconds is
Explanation
As given that, $x = (t - 2)^{2}$ velocity $v = dx/dt = d/dt(t - 2)^{2} = 2(t - 2)m/s$ $a = dv/dt = d/dt[2(t - 2)] = 2[1 - 0] = 2 m/s^{2} = 2 ms^{-2}$ at $t = 0$; $v_{0} = 2(0 - 2) = -4 m/s$ $t = 2 s$; $v_{2} = 2(2 - 2) = 0 m/s$ $t = 4 s$; $v_{4} = 2(4 - 2) = 4 m/s$ v - t graph is shown in diagram. Distance travelled = area between time axis of the graph = area OAC + are ABD = $\frac{1}{2}OA \times OC + \frac{1}{2}AD \times BD = 8 m$ If displacement occurs = $-\frac{1}{2} \times OA \times OC + \frac{1}{2} \times AD \times BD = 0$