Question

A body covers 10m in the second second and 25m in fifth second of its motion. If the motion is uniformly accelerated, how far will it go in seventh second?​

Answer 1

We know that

$S_{n^{th}}=u+\dfrac{a(2n-1)}{2}$

Where,

S is the displacement in nth second, a is the acceleration and n is the time and u is the initial velocity.

$\rule{200}{2}$

Displacement in second second:-

$S_2=u+\dfrac{a(4-1)}{2}\\ \\ \\S_2=u+\dfrac{3}{2}a\\ \\ \\ \dfrac{3}{2}a=10-u\\ \\ \\a=\dfrac{2(10-u)}{3}$

$\rule{200}{2}$

Displacement in Fifth second :-

$S_5=u+\dfrac{a(10-1)}{2}\\ \\ \\S_5=u+\dfrac{9a}{2}\\ \\ \\a=\dfrac{2}{9}(25-u)$

$\rule{200}{2}$

Since,body is uniformly accelerated.

Thus, Both of the accelerations must be equal.

Therefore :-

$\frac{2}{3}(10 - u) = \frac{2}{9}(25 - u) \\ \\ \\10 - u = \frac{(25 - u)}{3} \\ \\ \\3(10 - u) = 25 - u \\ \\ \\30 - 3u = 25 - u \\ \\ \\3u - u = 30 - 25 \\ \\ \\2u = 5 \\ \\ \\u = \frac{5}{2} \\ \\ \\u = 2.5$

$\rule{200}{2}$

Hence,

Initial velocity = 2.5 m/sec

Acceleration will be :-

$a = \frac{2}{3}(10 - u) \\ \\ \\a = \frac{2}{3}(10 - 2.5) \\ \\ \\a = \frac{2 \times 7.5}{3} \\ \\ \\a = 5$

Acceleration = 5 m/sec²

$\rule{200}{2}$

Displacement in 7th second will be :-

$S_7=u+\dfrac{a(2n-1)}{2}\\ \\ \\S_7=2.5+\dfrac{5*13}{2}\\ \\ \\S_7=2.5(1+13)\\ \\ \\S_7=2.5*14\\ \\ \\S_7=35\:\:m$

$\rule{200}{2}$

Answer 2

It cover 10 m in 2nd second of its Motion

We know

$s = ut + \frac{1}{2} at {}^{2}$

T =Time=2 seconds

S=10 m

$10 = u \times 2 + \frac{1}{2} a \times {2}^{2} \\ \\ 10 = 2u + 2a \\ \\ 10 = 2(u + a) \\ \\ u + a = \frac{10}{2} \\ \\ u + a = 5 \\ \\ a = 5 - u$

It cover 25 m in 5th second

We know

$s = ut + \frac{1}{2} {at}^{2}$

S=25 m

T= Time given=5 seconds

$25 = u \times 5 + \frac{1}{2} a \times {5}^{2} \\ \\ 25 = 5u + \frac{25a}{2} \\ \\ 25 = \frac{10u + 25a}{2} \\ \\ 25 \times 2 = 10u + 25a \\ \\ 50 = 5(2u + 5a) \\ \\ \frac{50}{5} = 2u + 5a \\ \\ 10 = 2u + 5a \\ \\ 10 - 2u = 5a \\ \\ a = \frac{10 - 2u}{5}$

Since it was uniformally accerlating

Accerlation in 2nd second=Accerlation in 5th second

$5 - u = \frac{10 - 2u}{5} \\ \\ 5(5 - u) = 10 - 2u \\ \\ 25 - 5u = 10 - 2u \\ \\ - 5u + 2u = 10 - 25 \\ \\ - 3u = - 15 \\ \\ u = 5$

Accerlation=5-u

=5-u

=0 m/s²

$s_{7} = ut + \frac{1}{2} at {}^{2} \\ \\ = 5 \times 7 + \frac{1}{2} \times 0 \times {7}^{2} \\ \\ 5 \times 7 \times 0 \\ \\ = 35$

Distance travelled in 7th second=35 m