Question

A body is moving along the +ve x axis with uniform acceleration of $- 4m/{s^2}$ . Its velocity at $x = 0$ is 10 m/s. the time taken by the body to reach a point at $x = 12m$ is
(A) (2s, 3s)
(B) (3s, 4s)
(C) (4s, 8s)
(D) (1s, 2s)

Answer 1

The acceleration being negative means that the acceleration is directed towards the negative x axis. The second equation of motion can be used to calculate the time.

Formula used: In this solution we will be using the following formulae;
$s = ut \pm \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered by a body undergoing acceleration, $u$ is the initial velocity, and $t$ is the time taken to cover the distance, and $a$ is the value of the acceleration of the body.

Complete Step-by-Step solution
A body is said to move along the positive x axis, and acceleration is defined to be equal to $- 4m/{s^2}$ . This shows that the acceleration is in the opposite direction as velocity (direction of motion), and thus in the direction of the negative x axis. Now the velocity was initialized at $x = 0$ to be equal to 10 m/s. we are to find the time taken to by the body to reach a point $x = 12m$ .
If observed, we have all the quantities needed to use the second equation of motion, given by
$s = ut \pm \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered by a body undergoing acceleration, $u$ is the initial velocity, and $t$ is the time taken to cover the distance, and $a$ is the value of the acceleration of the body.
Inserting all known values,
$12 = 10t - \dfrac{1}{2}\left( 4 \right){t^2}$
$\Rightarrow 12 = 10t - 2{t^2}$
Rearranging the equation and dividing through by 2, we have
${t^2} - 5t + 6 = 0$
Solving the quadratic equation which can be given as
$t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Hence, we have that
$t = \dfrac{{5 \pm \sqrt {{5^2} - 4\left( 6 \right)} }}{2}$
By computation,
$t = \dfrac{{5 \pm 1}}{2}$
Hence,
$\Rightarrow t = \dfrac{6}{2} = 3s{\text{ and }}t = \dfrac{4}{2} = 2s$
Hence, your correct answer is A.

Note
The presence of two answers signifies that when the object is at point $x = 12m$ , it could be at anyone of these times. This can occur also for example in a case of an object thrown upwards. At a particular point, the object will be there twice. When it is going up and when it is coming down which happens after different times.