Question

If the rate of change in velocity w.r.t time is constant and its position after 6th second will be tge same as that after 11th second then the particle returns to the starting point after

kindly give an explanation

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Answer 1

Rate of change of velocity w.r.t time is known as acceleration.

According to question rate of change in velocity w.r.t time is constant i.e. acceleration is constant.

As it is given that position after sixth second will be the same as that after eleventh second. So, by this it is clear that initial velocity (u) and acceleration (a) of particle are in opposite direction.

By using second equation of motion we can write the position at sixth & eleventh second.

Position of particle at sixth second:

$\rm \leadsto s_6 = ut_6 - \dfrac{1}{2}at_6^2 \\ \\ \rm \leadsto s_6 = 6u - \dfrac{1}{2}a {(6)}^{2} \\ \\ \rm \leadsto s_6 = 6u - \dfrac{1}{2}a \times 36 \\ \\ \rm \leadsto s_6 = 6u - 18a$

Position of particle at eleventh second:

$\rm \leadsto s_{11} = ut_{11} - \dfrac{1}{2}at_{11}^2 \\ \\ \rm \leadsto s_{11} = 11u - \dfrac{1}{2}a {(11)}^{2} \\ \\ \rm \leadsto s_{11} = 11u - \dfrac{121}{2}a$

From question we can write:

$\rm \leadsto s_6 = s_{11} \\ \\ \rm \leadsto 6u - 18a = 11u - \dfrac{121}{2} a \\ \\ \rm \leadsto 11u - 6u = \dfrac{121}{2} a - 18a \\ \\ \rm \leadsto 5u = \dfrac{121 - 36}{2} a \\ \\ \rm \leadsto 5u = \frac{85}{2} a \\ \\ \rm \leadsto u = \dfrac{17}{2} a$

When particle returns to its starting point it's displacement is zero.

So, by applying second equation of motion we get:

$\rm \leadsto s = ut - \dfrac{1}{2} a {t}^{2} \\ \\ \rm \leadsto 0 = \dfrac{17}{2} at - \dfrac{1}{2} a {t}^{2} \\ \\ \rm \leadsto \dfrac{1}{ \cancel{2}} \cancel{a} {t}^{ \cancel{2}} = \dfrac{17}{ \cancel{2}} \cancel{at} \\ \\ \rm \leadsto t = 17 \: s$

$\therefore$ Particle returns to the starting point after 17 seconds.

Answer 2

S6 = S11

6u - 1/2a×6² = 11u - 1/2a×11²

5u = 85a/2

a = 2/17u

s = ut - 1/2at²

0 = ut - 1/2 × 2/17ut²

t/17 = 1

t = 17 s