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  • The instantaneous velocity of a particle moving in a straight line is given as <img alt="\mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cmathrm%7Bv%7D%3D%5Calpha%20%5Cmathrm%7Bt%7D&plus;%5Cbeta%

The instantaneous velocity of a particle moving in a straight line is given as \mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}, where \alpha$ and $\beta are constants. The distance travelled by the particle between 1 \mathrm{~s}$ and $2 \mathrm{~s} is :
 
Option: 1 3 \alpha+7 \beta
Option: 2 \frac{3}{2} \alpha+\frac{7}{3} \beta
Option: 3 \frac{\alpha}{2}+\frac{\beta}{3}
Option: 4 \frac{3}{2} \alpha+\frac{7}{2} \beta

Answers (1)

best_answer

v= \alpha t+\beta t^{2}
\frac{ds}{dt}= \alpha t+\beta t^{2}
\int ds=\int \alpha t\, dt+\int \beta t^{2}\, dt
s= \alpha \left [ \frac{t^{2}}{2} \right ]^{2}_{1}+\beta \left [ \frac{t^{3}}{2} \right ]^{2}_{1}
s= \alpha \left [ \frac{4}{2}-\frac{1}{2} \right ]+\beta \left [ \frac{8}{3}-\frac{1}{3} \right ]
s= \frac{3}{2}\alpha +\frac{7}{3}\beta
The correct option is (2)

Posted by

vishal kumar

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