If the velocity of a particle is $v = At + Bt^2$, where $A$ and $B$ are constants, then the distance travelled

Question

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

Accepted Answer

Velocity is defined as the rate of change of position with respect to time ($v = \frac{dx}{dt}$) . To find the distance travelled between two instants, we integrate the velocity function over the given time interval.

1.  **Set up the integral:**
    $$ 	ext{Distance} = \int_{t_1}^{t_2} v \, dt = \int_{1}^{2} (At + Bt^2) \, dt $$

2.  **Integrate:**
    Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1}$:
    $$ x = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} ight]_{1}^{2} $$

3.  **Apply limits (Upper limit 2, Lower limit 1):**
    $$ x = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} ight) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} ight) $$
    $$ x = \left( \frac{4A}{2} + \frac{8B}{3} ight) - \left( \frac{A}{2} + \frac{B}{3} ight) $$
    $$ x = 2A + \frac{8B}{3} - \frac{A}{2} - \frac{B}{3} $$

4.  **Simplify:**
    $$ x = \left( 2A - \frac{A}{2} ight) + \left( \frac{8B}{3} - \frac{B}{3} ight) $$
    $$ x = \frac{3A}{2} + \frac{7B}{3} $$

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