The position-time (x-t) graph of a moving bod | vedclass

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(B) Given, mass of the body, $m = 2 \, kg$.Impulse is defined as the change in momentum: $	ext{Impulse} = \Delta p = p_f - p_i = m(v_f - v_i)$.From t

Vedclass · Accepted Answer

$-1.5 \, kg \cdot m/s$

Vedclass · Answer

(B) Given, mass of the body, $m = 2 \, kg$.Impulse is defined as the change in momentum: $	ext{Impulse} = \Delta p = p_f - p_i = m(v_f - v_i)$.From the position-time graph, the velocity is the slope of the $x-t$ graph $(v = \frac{dx}{dt})$.For $t $v_i = \frac{3 - 0}{4 - 0} = 0.75 \, m/s$.For $t > 4 \, s$, the velocity $v_f$ is the slope of the horizontal line:$v_f = 0 \, m/s$.Therefore, the impulse at $t = 4 \, s$ is:$	ext{Impulse} = m(v_f - v_i) = 2 \, kg 	imes (0 - 0.75 \, m/s) = -1.5 \, kg \cdot m/s$.Thus, the correct option is $B$.

The position-time $(x-t)$ graph of a moving body of mass $2 \, kg$ is shown in the figure. The impulse on the body at $t=4 \, s$ is

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