A body leaving a certain point O moves with an acceleration which is constant. At the end of the first second after it left O, its velocity is $1.5 ms^{-1}$. At the end of the sixth second after it left O the body stops momentarily and begins to move backwards. The distance travelled by the body before it stops momentarily is
a) $9 m$
b) $18 m$
c) $5.4 m$
d) $36 m$

Hint : First we will make the required diagram according to the given question. We will use the first equation of motion to find the initial velocity and the acceleration of the body. After that we will find the distance covered by the body using the third equation of motion.

Complete step-by-step solution:
Here, the required diagram is:

We will use the first equation of motion.
$v = u + at$
where, v is the final velocity.
u is the initial velocity.
a is acceleration.
t is time taken.
Velocity after one second will be $1.5 ms^{-1}$.
We will apply the first equation of motion.
$1.5 = u + 1a$ ……($1$)
Velocity after six seconds will be $0 ms^{-1}$.
We will apply the first equation of motion.
$0 = u + 6a$
$\implies u = -6a$ ……($2$)
Put ($2$) in ($1$), we get
$1.5 = -6a + 1a$
$\implies 1.5 = -5a$
We get acceleration,
$a = -0.3 m s^{-2}$
Now, we will find the initial velocity by equation ($2$),
$u = -6 \times – 0.3 = 1.8 m s^{-1}$
Now, we have to find the distance using the third equation of motion.
$v^{2} – u^{2} = 2as$
$\implies 0^{2} – 1.8^{2} = 2\times -0.3 \times s$
$\implies s = 5.4 m$
The distance travelled by the body before it stops momentarily is $5.4 m$.
Option (c) is correct.

Note: Equations of motion are equalizations that explain the behavior of a dynamic system in terms of its action as a function of time. More particularly, the equations of motion explain the behavior of a physical system as a collection of mathematical functions in terms of changing variables.