The impulse due to the force F→=ai^+btj^, where a=2 and b=4N/s if this force acts from ti=0 to tf=0.3s is

J=∫titfF→dt=∫00.3(ai^+btj^)dt J=a∫00.3dti^+b∫00.3t dt j^=a[t]00.3i^+b[t22]00.3j⌢ =2×0.3×i^+4×(0.3)22=0.6i^+0.18j^ N.sec

The impulse due to the force F→=ai^+btj^, where a=2 and b=4N/s if this force acts from ti=0 to tf=0.3s is

J=∫titfF→dt=∫00.3(ai^+btj^)dt J=a∫00.3dti^+b∫00.3t dt j^=a[t]00.3i^+b[t22]00.3j⌢ =2×0.3×i^+4×(0.3)22=0.6i^+0.18j^ N.sec

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The impulse due to the force $\vec{F} = a \hat{i} + b t \hat{j},$ where $a = 2 a n d b = 4 N / s$ if this force acts from $t_{i} = 0 t o t_{f} = 0.3 s i s$

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$0.6 \hat{i}$

$0.6 \hat{i} + 0.18 \hat{j}$

$0.18 \hat{j}$

answer is B.

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Detailed Solution

$J = \int_{t_{i}}^{t_{f}} \vec{F} d t = \int_{0}^{0.3} (a \hat{i} + b t \hat{j}) d t J = a \int_{0}^{0.3} d t \hat{i} + b \int_{0}^{0.3} t d t \hat{j} = a {[t]}_{0}^{0.3} \hat{i} + b {[\frac{t^{2}}{2}]}_{0}^{0.3} \overset{⌢}{j} = 2 \times 0.3 \times \hat{i} + 4 \times \frac{{(0.3)}^{2}}{2} = 0.6 \hat{i} + 0.18 \hat{j} N . \sec$