Question

An object moving with $6 ms^{-1}$ executes an acceleration $2 ms^{-2}$ in the next $3$ seconds. How much distance did it cover?

Answer 1

In the question, initial velocity, acceleration and time taken are given. First, we will find the final velocity of the object using the first equation of motion. Then, we will use that final velocity for finding the distance covered by the object using the third equation of motion.

Complete step by step answer:
Given: initial velocity of an object, $u = 6 ms^{-1}$
acceleration of an object, $a = 2 ms^{-2}$
time taken by the object, $t = 3 seconds$
Using Newton’s first equation of motion,
$v = u + at$
Put all the values in the above equation.
$\implies v = 6 + 2 \times 3$
$\implies v = 12 ms^{-1}$
Applying Newton’s third equation of motion,
$v^{2} – u^{2} = 2as$
Put all the values in the above equation.
$\implies 12^{2} – 6^{2} = 2 \times 2 \times s$
$\implies 4s = 144 - 36$
$\implies s = \dfrac{108}{4} = 27 m$
$27 m$ is the distance covered by the object.

Note: We can find distance covered by an object using Newton’s second equation of motion,
$S = ut + \dfrac{1}{2} a t^{2}$
Put all the values in the above equation.
$\implies S = 6 \times 3 + \dfrac{1}{2} \times 2 \times 3^{2}$
$\implies S = 6 \times 3 + \dfrac{1}{2} \times 2 \times 9$
$\implies S = 18 + 9$
$\implies S = 27 m$
$27 m$ is the distance covered by the object.