Question

A particle travels 10m in the first 5s and 10 m in the next 3s. Assuming constant acceleration what is the distance travelled in next 2s.
$\left( A \right)8.3m$
$\left( B \right)9.3m$
$\left( C \right)10.3m$
$\left( D \right)$ None of above

Answer 1

Hint: In this question use the equation of second law of motion which is given as $s = ut + \dfrac{1}{2}a{t^2}$and later on in the solution use the concept that the distance travelled in the next seconds is the sum of previous distance and present distance similarly for time also so use these properties to reach the solution of the question.

Formula used – $s = ut + \dfrac{1}{2}a{t^2}$

Complete Step-by-Step solution:
Given data:
${s_1} = 10m$, ${t_1} = 5s$
So from second equation of motion which is given as:
$s = ut + \dfrac{1}{2}a{t^2}$................. (1), where s = distance travelled.
                                                                  u = initial velocity.
                                                                  a = acceleration.
                                                                  t = time.
Therefore for 1st case we have,
${s_1} = u{t_1} + \dfrac{1}{2}at_1^2$
Now substitute the values we have,
$ \Rightarrow 10 = 5u + \dfrac{{{5^2}}}{2}a$..................... (2)
Now in the next 3s particles travel 10m.
So the total distance particle covered ${s_2} = 10 + 10 = 20m$ and the total time taken ${t_2} = 5 + 3 = 8s$
So from equation (1) we have,
$ \Rightarrow {s_2} = u{t_2} + \dfrac{1}{2}at_2^2$
Now substitute the values we have,
$ \Rightarrow 20 = 8u + \dfrac{{{8^2}}}{2}a$
$ \Rightarrow a = \dfrac{{40 - 16u}}{{64}}$................... (3)
Now substitute this value in equation (2) we have,
$ \Rightarrow 10 = 5u + \dfrac{{{5^2}}}{2}\left( {\dfrac{{40 - 16u}}{{64}}} \right)$
Now simplify this equation we have,
$ \Rightarrow 10 = 5u + \left( {\dfrac{{1000 - 400u}}{{128}}} \right)$
$ \Rightarrow 1280 - 640u = 1000 - 400u$
$ \Rightarrow 240u = 280$
$ \Rightarrow u = \dfrac{{280}}{{240}} = \dfrac{7}{6}$ m/s.
Now from equation (3) we have,
$ \Rightarrow a = \dfrac{{40 - 16u}}{{64}} = \dfrac{{40 - 16\left( {\dfrac{7}{6}} \right)}}{{64}} = \dfrac{{40 - \dfrac{{56}}{3}}}{{64}} = \dfrac{{120 - 56}}{{64 \times 3}}{\text{ = }}\dfrac{{64}}{{64 \times 3}}{\text{ = }}\dfrac{1}{3}{\text{m/}}{{\text{s}}^{\text{2}}}$
Now we have to find out the distance travelled in the next 2s.
So the total distance particle covered ${s_3}$ and the total time taken ${t_3} = 8 + 2 = 10s$
So from equation (1) we have,
$ \Rightarrow {s_3} = u{t_3} + \dfrac{1}{2}at_3^2$
As acceleration is constant so there is no change.
So substitute the values we have,
$ \Rightarrow {s_3} = \dfrac{7}{6}\left( {10} \right) + \dfrac{1}{2}\left( {\dfrac{1}{3}} \right){\left( {10} \right)^2} = \dfrac{{70 + 100}}{6} = \dfrac{{170}}{6} = \dfrac{{85}}{3}m$
Now the distance (s) travelled in next 2 sec is
$s = {s_3} - {s_2} = \dfrac{{85}}{3} - 20 = \dfrac{{25}}{3} = 8.3m$
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of questions the key concept is equation of second law of motion so just substitute the given values in this equation for different cases and calculate the value of initial velocity and acceleration as above using any method of solving linear equations (such as substitution, elimination, cross multiplication etc.), here we use substitution method, then calculate the distance covered by the particle in next 2s using these values as acceleration is constant we will get the required answer.