Displacement of a particle of mass 2, kg movi | vedclass

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(B) Given,mass $m = 2\, kg$ and displacement $s = (2t^3 + 2)\, m$.Velocity $v$ is the rate of change of displacement with respect to time:$v = \frac{d

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Given,mass $m = 2\, kg$ and displacement $s = (2t^3 + 2)\, m$.Velocity $v$ is the rate of change of displacement with respect to time:$v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 + 2) = 6t^2\, m/s$.At $t = 0\, s$,initial velocity $v_i = 6(0)^2 = 0\, m/s$.At $t = 1\, s$,final velocity $v_f = 6(1)^2 = 6\, m/s$.Impulse $J$ is equal to the change in linear momentum $\Delta P$:$J = \Delta P = m(v_f - v_i)$.Substituting the values:$J = 2\, kg 	imes (6\, m/s - 0\, m/s) = 2 	imes 6 = 12\, N-s$.

Displacement of a particle of mass $2\, kg$ moving in a straight line varies with time as $s = (2t^3 + 2)\, m$. The impulse of the force acting on the particle over a time interval between $t = 0$ and $t = 1\, s$ is .......... $N-s$.

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