$A$ girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in the figure. Plot a velocity-time graph for the same.
(N/A) $1$. From the displacement-time graph,the velocity is the slope of the line.
$2$. For the interval $0-50 \, s$,the displacement changes from $0 \, m$ to $100 \, m$. Thus,velocity $v_1 = \frac{100 \, m - 0 \, m}{50 \, s - 0 \, s} = 2 \, m/s$.
$3$. For the interval $50-100 \, s$,the displacement changes from $100 \, m$ to $0 \, m$. Thus,velocity $v_2 = \frac{0 \, m - 100 \, m}{100 \, s - 50 \, s} = \frac{-100 \, m}{50 \, s} = -2 \, m/s$.
$4$. The velocity-time graph will show a constant velocity of $2 \, m/s$ from $0$ to $50 \, s$ and a constant velocity of $-2 \, m/s$ from $50$ to $100 \, s$.
$2$. For the interval $0-50 \, s$,the displacement changes from $0 \, m$ to $100 \, m$. Thus,velocity $v_1 = \frac{100 \, m - 0 \, m}{50 \, s - 0 \, s} = 2 \, m/s$.
$3$. For the interval $50-100 \, s$,the displacement changes from $100 \, m$ to $0 \, m$. Thus,velocity $v_2 = \frac{0 \, m - 100 \, m}{100 \, s - 50 \, s} = \frac{-100 \, m}{50 \, s} = -2 \, m/s$.
$4$. The velocity-time graph will show a constant velocity of $2 \, m/s$ from $0$ to $50 \, s$ and a constant velocity of $-2 \, m/s$ from $50$ to $100 \, s$.
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