Question

If the velocity of a particle is $v=A t+B t^{2}$, where $A$ and $B$ are constant, then the distance travelled by it between 1 s and 2 s is:

A. $\frac{3}{2} A+4 B$
B. $3 A+7 B$
C. $\frac{3}{2} A+\frac{7}{3} B$
D. $\frac{A}{2}+\frac{B}{3}$

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Answer

Given that the velocity of a particle is $v = A t + B t^{2}$, where $A$ and $B$ are constants, we need to determine the distance travelled by the particle between $1$ second and $2$ seconds.

:

  1. Formulating the problem: The velocity equation is given by: $$ v = A t + B t^{2} $$ Velocity $v$ can also be expressed as the first derivative of the position $x$ with respect to time $t$: $$ v = \frac{dx}{dt} $$ Thus, $$ \frac{dx}{dt} = A t + B t^{2} $$

  2. Rearranging and integrating: Separating the variables, we get: $$ dx = (A t + B t^{2}) , dt $$ To find the distance travelled between $1$ second and $2$ seconds, integrate both sides with respect to $t$ from $1$ to $2$: $$ \int_{1}^{2} dx = \int_{1}^{2} (A t + B t^{2}) , dt $$ On the left side, integrating $dx$ gives the change in position $x$: $$ \Delta x = \int_{1}^{2} (A t + B t^{2}) , dt $$

  3. Evaluating the integral: Split the integral into two parts: $$ \Delta x = A \int_{1}^{2} t , dt + B \int_{1}^{2} t^{2} , dt $$ Calculate each integral separately: [ \int_{1}^{2} t , dt = \left[ \frac{t^{2}}{2} \right]{1}^{2} = \left( \frac{2^{2}}{2} - \frac{1^{2}}{2} \right) = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} ] [ \int{1}^{2} t^{2} , dt = \left[ \frac{t^{3}}{3} \right]_{1}^{2} = \left( \frac{2^{3}}{3} - \frac{1^{3}}{3} \right) = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} ]

  4. Combining the results: Substitute these back into the equation for $\Delta x$: $$ \Delta x = A \left(\frac{3}{2}\right) + B \left(\frac{7}{3}\right) $$

So, the distance travelled by the particle between $1$ second and $2$ seconds is: $$ \Delta x = \frac{3}{2} A + \frac{7}{3} B $$

Conclusion:

The correct answer is: C. $\frac{3}{2} A + \frac{7}{3} B$


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