Question

The position of a body moving along the x axis as a function of time t is given as $x = {t^2} - 2t$metre. The distance travelled by the body in the first two seconds is
$1){\text{ Zero}}$
$2){\text{ 1m}}$
$3){\text{ 2m}}$
$4){\text{ 4m}}$

Answer 1

Hint: In order to solve we will use the distance time relation given the question to find the find the position of a body moving along the x axis first of we will find the distance from zero second to one second then we will find the distance between one second to two second and then we will sum both the distance to find the total distance

Complete step-by-step solution:
We are given the distance time relation that is
$x = {t^2} - 2t$
In order to find the total distance object has travelled we find the distance at different interval of time
First we will find distance from zero second to one second
Distance from $t = 0$ to $t = 1$
We will insert the value of t in the given equation
When t is equal to 0 second
$x = {t^2} - 2t$
Substituting value of t
${x_0} = {0^2} - 2(0)$
${x_0} = 0$
Therefore, at $t = 0$ ${x_0} = 0$
When t is equal to 1 second
${x_1} = {t^2} - 2t$
Substituting value of t
${x_1} = {1^2} - 2(1)$
${x_1} = - 1$
${x_1}$ cannot be negative as distance is the magnitude so ${x_1} = 1m$
Therefore, at $t = 1$ ${x_1} = 1m$
Distance travelled in from $t = 0$ to $t = 1$ is $1m$
Distance from $t = 1$ to $t = 2$
We will insert the value of t in the given equation
When t is equal to 1 second
${x_1} = {t^2} - 2t$
Substituting value of t
${x_1} = {1^2} - 2(1)$
${x_1} = - 1$
${x_1}$ cannot be negative as distance is the magnitude so ${x_1} = 1m$
Therefore, at $t = 1$ ${x_1} = 1m$
When t is equal to 2 second
${x_2} = {t^2} - 2t$
Substituting value of t
${x_2} = {2^2} - 2(2)$
${x_2} = 4 - 4$
${x_2} = 0$
${x_2} = {0^2} - 2(0)$
${x_2} = 0$
Therefore, at $t = 2$ ${x_2} = 0$
Distance travelled from $t = 1$ to $t = 2$ is $1m$
Now adding distance travelled from $t = 0$ to $t = 1$ and the distance travelled from $t = 1$ to $t = 2$
$t = 0$ to $t = 1$ is $1m$
 $t = 1$ to $t = 2$ is $1m$
Adding
$ \Rightarrow 1m + 1m$
$ \Rightarrow 2m$
Hence the total distance is $2m$ and the correct option is $3){\text{ 2m}}$.

Note: Many of the people will make mistakes by not removing negative signs from the distance but distance is scalar quantity it is only the magnitude and magnitude can never be negative distance can only increase it can never be decreased so in above question we will take distance positive where it comes out to be negative.