Question

If the velocity of a particle is $v = at + b{t^2}$ where $a$ and $b$ are constant then the distance travelled by it between 1s and 2s is
(A) $3A + 7B$
(B) $\dfrac{3}{2}A + \dfrac{7}{3}B$
(C) $\dfrac{A}{2} + \dfrac{B}{3}$
(D) $\dfrac{3}{2}A + 4B$

Answer 1

Hint: A body is said to be in a state of motion if its position changes with respect to time. If a body moves by covering equal distance in equal intervals of time is called uniform motion.

Complete Step By Step Solution:
We know that,
$\dfrac{{dx}}{{dt}} = v$ where $x$ is the distance travelled
By cross multiplying we get,
$dx = vdt$ where $v = at + b{t^2}$
In order to find the distance, we have to integrate on both sides
$\int\limits_0^x {dx} = \int\limits_{t = 1}^{t = 2} {vdt} $
We get;
$x = \int\limits_{t = 1}^{t = 2} {(At + B{t^2})dt} $
    $ = \int\limits_{t = 1}^{t = 2} {Atdt + \int\limits_{t = 1}^{t = 2} {B{t^2}dt} } $
    $ = {\left[ {[A\dfrac{{{t^2}}}{2}] + [B\dfrac{{{t^3}}}{3}]} \right]_1}^2$
    $ = \left[ {[\dfrac{A}{2}({2^2} - {1^2}) + \dfrac{B}{3}({2^3} - {1^3})} \right]$
    $ = \left[ {[\dfrac{A}{2}(4 - 1) + \dfrac{B}{3}(8 - 1)} \right]$
    $ = \left[ {[\dfrac{A}{2}(3) + \dfrac{B}{3}(7)} \right]$
    $x = 3\dfrac{A}{2} + 7\dfrac{B}{3}$

Additional information:
Velocity --Velocity is defined as the rate of change of position of an object with respect to a frame of reference which is a function of time.
Instantaneous velocity -Instantaneous velocity is defined as the rate of change of displacement.
Acceleration--Acceleration is defined as the rate of change of velocity of an object with respect to time. It is a vector quantity.
Displacement-Displacement refers to the shortest distance covered by an object between the initial and final point.

Note:
The three equations of motions are
$
  v = u + at \\
  s = ut + \dfrac{1}{2}a{t^2} \\
  {v^2} = {u^2} + 2as \\
$