At any instant the velocity of a particle of mass 500 g is (2t i^+3t2j^)ms−1. If the force acting on the particle at t=1s is ( i^+xj^)N. Then the value of x will be

m=0.5kg,v¯=2i^+3t2j^a¯=dvdt=2i^+6tj^F¯=ma¯=0.5=(2i^+6tj^)=i^+3t j^t=1s,F¯=i^+3j^x=3N

At any instant the velocity of a particle of mass 500 g is (2t i^+3t2j^)ms−1. If the force acting on the particle at t=1s is ( i^+xj^)N. Then the value of x will be

m=0.5kg,v¯=2i^+3t2j^a¯=dvdt=2i^+6tj^F¯=ma¯=0.5=(2i^+6tj^)=i^+3t j^t=1s,F¯=i^+3j^x=3N

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At any instant the velocity of a particle of mass 500 g is $(2 t \hat{i} + 3 t^{2} \hat{j}) {ms}^{- 1}$ . If the force acting on the particle at $t = 1 s$ is $(\hat{i} + x \hat{j}) N$ . Then the value of x will be

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Detailed Solution

$\begin{array}{l} m = 0.5 kg, \bar{v} = 2 \hat{i} + 3 t^{2} \hat{j} \\ \bar{a} = \frac{dv}{dt} = 2 \hat{i} + 6 t \hat{j} \\ \bar{F} = m \bar{a} = 0.5 = (2 \hat{i} + 6 t \hat{j}) = \hat{i} + 3 \overset{}{t} \hat{j} \\ t = 1 s, \bar{F} = \hat{i} + 3 \hat{j} \\ x = 3 N \end{array}$