At time t=0, a particle leaves the origin and | vedclass

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(B) Given, $v(t) = v_0(1 - t/t_0)$ with $v_0 = 10 \ m/s$ and $t_0 = 10 \ s$.The velocity becomes zero at time $t_1$ when $1 - t_1/t_0 = 0$, so $t_1 =

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(B) Given, $v(t) = v_0(1 - t/t_0)$ with $v_0 = 10 \ m/s$ and $t_0 = 10 \ s$.The velocity becomes zero at time $t_1$ when $1 - t_1/t_0 = 0$, so $t_1 = t_0 = 10 \ s$.For $0 \le t \le 10 \ s$, the particle moves in the positive direction. The displacement $s_1$ is given by the integral of velocity:$s_1 = \int_0^{10} v_0(1 - t/t_0) dt = v_0 [t - t^2/(2t_0)]_0^{10} = 10 [10 - 100/20] = 10 [10 - 5] = 50 \ m$.For $10 \le t \le 20 \ s$, the velocity becomes negative, meaning the particle moves in the negative direction. The displacement $s_2$ is:$s_2 = \int_{10}^{20} v_0(1 - t/t_0) dt = 10 [t - t^2/20]_{10}^{20} = 10 [(20 - 400/20) - (10 - 100/20)] = 10 [(20 - 20) - (10 - 5)] = 10 [0 - 5] = -50 \ m$.The total distance covered is the sum of the magnitudes of displacements: $d = |s_1| + |s_2| = |50| + |-50| = 100 \ m$.

At time $t=0$, a particle leaves the origin and moves in the positive direction of the $X$-axis. If the velocity of the particle varies as $v(t)=v_0(1-t/t_0)$, where $|v_0|=10 \ m/s$ and $t_0=10 \ s$, then the distance covered by the particle during the first $20 \ s$ is: (in $m$)

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