Question

The position of a body moving along the x−axis at time t is given by $x = {t^2} - 4t + 6$m. The distance travelled by body in time interval t=0 to t=3 s is
A) 5 m
B) 7 m
C) 4 m
D) 3 m

Answer 1

When position of a body is given in terms of the time then position at any particular time can be known by just putting the value of time in that relation. Again the velocity of the body can be determined by taking derivatives of the position function w.r.t. time. Total distance travelled by the body is given by the length of the total path it has traversed throughout the timespan.

Formula used:
Velocity of a body can be obtained from the position as:
$v = \dfrac{{dx}}{{dt}}$ ……………………...(1)
Where,
v is the velocity of the body,
$\dfrac{d}{{dt}}$is the derivative w.r.t time t,
x is the position of the body.

Complete step by step answer:
Given: Position of the body moving in x-axis is given by $x = {t^2} - 4t + 6$m.

To find: Total distance travelled within t=0 to t=3 s.

Step 1
First, use the position function x in eq.(1) to get the expression for velocity v as:

v = \dfrac{d}{{dt}}({t^2} - 4t + 6) \\\ \therefore v = 2t - 4 \\\ $$............................(2) Step 2 Equate the expression of eq.(2) with 0 to get when v changes its sign:

v = 2t - 4 = 0 \\
\Rightarrow 2t = 4 \\
\therefore t = 2 \\

So, at t=2s velocity of the body becomes 0. For t<2 we have, v<0 and for t>2 we have v>0. Hence, the velocity changes its sign at t=2s. So, movement direction of the body will also change at t=2s. So, first we need to find the distance travelled by the body in two parts, first from 0 to 2 s then from 2 to 3 s. Step 3 Now, put t=0s , t=2s, t=3s respectively in the given expression to get the position of body at each time as: $ {x_{t = 0}} = {0^2} - 4 \times 0 + 6 = 6m \\\ {x_{t = 2}} = {2^2} - 4 \times 2 + 6 = 2m \\\ {x_{t = 3}} = {3^2} - 4 \times 3 + 6 = 3m \\\ $.............................(3) Step 4 Use eq.(3) to get the distance travelled between 0 to 2 s as: ${d_{0 \to 2}} = \left| {{x_{t = 2}} - {x_{t = 0}}} \right| = \left| {2 - 6} \right|m = 4m$ Similarly, from eq.(3) get the distance travelled between 2 to 3s as: ${d_{2 \to 3}} = \left| {{x_{t = 3}} - {x_{t = 2}}} \right| = \left| {3 - 2} \right|m = 1m$ Hence, total distance travelled by the body between 0 to 3 s is given by: ${d_{0 \to 3}} = {d_{0 \to 2}} + {d_{2 \to 3}} = 4m + 1m = 5\,m$ **Correct answer:** Total distance travelled within 0 to 3 s is (a) 5 m. **Note:** Many students make mistakes while calculating total distance travelled by the body. They just take the final position at t=3 s and initial position at t= 0s and take their difference to get the answer. But that will give the displacement value, not total distance travelled. So, always be careful whether displacement or the total distance travelled has been asked.