Question

Please solve this question number 4

Answer 1

Answer:

$\boxed{\mathfrak{(4) \ \dfrac{3}{2} A+ \dfrac{7 }{3} B }}$

Explanation:

Relation of velocity of particle with time is given as:

$\rm v = At + Bt^2$

Rate of change of displacement is equal to velocity i.e.

$\rm v = \dfrac{dx}{dt}$

So,

$\rm \implies \dfrac{dx}{dt} = At + Bt^2 \\ \\ \rm \implies \int\limits^x_0dx = \int\limits^{t_2}_{t_1}(At + Bt^2).dt$

For the given question:

$\rm t_1 = 1 \ s$

$\rm t_2 = 2 \ s$

$\rm \implies \int\limits^x_0dx = \int\limits^{2}_{1}(At + Bt^2).dt \\ \\ \rm \implies x\Big|_0^x = \dfrac{A {t}^{2} }{2} + \dfrac{B {t}^{3} }{3} \Big|_1^2 \\ \\ \rm \implies x = \dfrac{A \times {2}^{2} }{2} + \dfrac{B \times {2}^{3} }{3} - (\dfrac{A \times {1}^{2} }{2} + \dfrac{B \times {1}^{3} }{3} )\\ \\ \rm \implies x = \dfrac{4A }{2} + \dfrac{8B }{3} - (\dfrac{A }{2} + \dfrac{B }{3} )\\ \\ \rm \implies x = \dfrac{4A }{2} + \dfrac{8B }{3} - \dfrac{A }{2} - \dfrac{B }{3} \\ \\ \rm \implies x = \dfrac{4A - A }{2} + \dfrac{8B - B}{3} \\ \\ \rm \implies x = \dfrac{3A }{2} + \dfrac{7B }{3}$