Question

An object is moving with uniform acceleration. Its velocity after \[4\,{\text{s}}\] is \[{\text{20 m}}{{\text{s}}^{ - 1}}\] and after \[7{\text{seconds}}\] is \[{\text{29 m}}{{\text{s}}^{ - 1}}\]. Find the distance travelled by the object in \[10{\text{th}}\] second.

Answer 1

Hint: We are asked to calculate the distance travelled by the object in \[10{\text{th}}\] second. To calculate this, recall the formula to find the distance covered by a body in \[n{\text{th}}\] second. Then find the value of acceleration of the object and use the first equation of motion to find its initial velocity. Put these values in the formula to get the required answer.

Complete step by step answer:
Given, after time \[{t_1} = 4\,{\text{s}}\] the velocity is \[{v_1} = 20{\text{ m}}{{\text{s}}^{ - 1}}\].After time \[{t_2} = 7{\text{s}}\] the velocity is \[{v_2} = 29{\text{ m}}{{\text{s}}^{ - 1}}\].
We are asked to find the distance travelled by the object in \[10{\text{th}}\] second.
The formula to find the distance covered by a body in \[n{\text{th}}\] second is written as,
\[{s_n} = u + \dfrac{1}{2}a\left( {2n - 1} \right)\]
where \[u\] is the initial velocity and \[a\] is the acceleration of the body.
At first we will find the acceleration of the object.
Acceleration can be defined as change in velocity per unit time, mathematically we write
\[a = \dfrac{{\Delta v}}{{\Delta t}}\] (i)
where \[\Delta v\] is the change in velocity and \[t\] is the time taken.

Here, change in velocity is,
\[\Delta v = {v_2} - {v_1}\]
Putting the values of \[{v_2}\] and \[{v_1}\] we get,
\[\Delta v = 29 - 20\]
\[ \Rightarrow \Delta v = 9\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}}\]
The time taken will be,
\[\Delta t = {t_2} - {t_1}\]
Putting the values of \[{t_2}\] and \[{t_1}\] we get,
\[\Delta t = 7 - 4\]
\[ \Rightarrow \Delta t = 3\,{\text{s}}\]
Now, putting the values of \[\Delta v\] and \[\Delta t\] in equation (i) we get the acceleration as,
\[a = \dfrac{9}{3}\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}\]
\[ \Rightarrow a = 3\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}\]

Now, to find the distance travelled by the object in \[10{\text{th}}\] second we need to find the initial velocity of the object. For this, we will use first equation of motion, which says
\[v = u + at\] (ii)
where \[u\] is the initial velocity, \[v\] is the final velocity, \[a\] is the acceleration and \[t\] is the time taken by a body.
Here, if we taken initial velocity as \[u\] and final velocity as \[v = {v_1} = 20{\text{ m}}{{\text{s}}^{ - 1}}\] then time taken is \[t = {t_1} = 4\,{\text{s}}\] and we got acceleration as \[a = 3\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}\]. Putting these values in equation (ii) we get,
\[20 = u + 3 \times 4\]
\[ \Rightarrow 20 = u + 12\]
\[ \Rightarrow u = 20 - 12\]
\[ \Rightarrow u = 8\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}\]
The formula to find the distance covered in \[n{\text{th}}\] second is given by,
\[{s_n} = u + \dfrac{1}{2}a\left( {2n - 1} \right)\] (iii)
Here, \[n = 10\]
Now, putting the values of \[u\], \[a\] and \[n\] in equation (iii) we get,
\[{s_n} = 8 + \dfrac{1}{2} \times 3\left( {2 \times 10 - 1} \right)\]
\[ \Rightarrow {s_n} = 8 + \dfrac{1}{2} \times 3 \times 19\]
\[ \Rightarrow {s_n} = 8 + 28.5\]
\[ \therefore {s_n} = 36.5\,{\text{m}}\]

Therefore, distance travelled by the object in \[10{\text{th}}\] second is \[36.5\,{\text{m}}\].

Note: Most of the time students get confused between the terms displacement and distance covered by a body. Displacement is the shortest distance between the initial and the final position and distance covered is the total length travelled by the body. If the initial and final positions are the same, then the displacement is zero but distance covered may be non zero.