a particle experiences constant acceleration for 20 second after start

1.	a particle experiences constant acceleration for 20 second after starting moving from rest. if it travels a distance s1 in the first 10 second and distance s2 in the next 10 second. find the relation between s1 and s2.
Answer» ANSWER: $\textsf{$S_2$ will be three times of $S_1$}$ GIVEN Constant acceleration for 20 s Starting moving from rest. $S_1$ in the first 10 seconds $S_2$ in the LAST 10 seconds TO FIND: $S_1$ , $S_2$ Relation between $S_1$ and $S_2$ EXPLANATION: $S_1$ in the first 10 seconds: $\huge{ \boxed{ \bf{ \sf{v = u + at}}}}$ u = 0 a = a m/s² t = 10 s v = 0 + a(10) v = 10a m/s $\huge{ \boxed{ \bf{ \sf{S = ut + \dfrac{1}{2}at^2}}}}$ $S_1 = 0(t) + \dfrac{1}{2}a(10)^2 \\ \\ \\ S_1 = \frac{100a}{2} \\ \\ \\ S_1 = 50a \: m$ $S_2$ in the last 10 seconds: After 10s the PARTICLE will have the velocity of v = at which is the final velocity for the first ten seconds and INITIAL velocity for the next ten seconds. $\huge{ \boxed{ \bf{ \sf{S = ut + \dfrac{1}{2}at^2}}}}$ u = at m/s a = a m/s² t = 10 s $S_2 = at(t) + \dfrac{1}{2}at^2 \\ \\ \\ S_2 =at^2\bigg(1 + \dfrac{1}{2}\bigg) \\ \\ \\ S_2 =a(100)(3/2) \\ \\ \\ S_2 = 150 a \: m\\ \\ \\S_1 = 50a\:\:metres\\ \\ \\S_2 = 150a\:\: metres\\ \\ \\S_2 = 3S_1$ ${\large{\bf\textsf{Hence $S_2$ will be three times of $S_1$}}}$

a particle experiences constant acceleration for 20 second after starting moving from rest. if it travels a distance s1 in the first 10 second and distance s2 in the next 10 second. find the relation between s1 and s2.

Answer»

ANSWER:

$\textsf{$S_2$ will be three times of $S_1$}$

GIVEN

Constant acceleration for 20 s

Starting moving from rest.

$S_1$ in the first 10 seconds

$S_2$ in the LAST 10 seconds

TO FIND:

$S_1$ , $S_2$

Relation between $S_1$ and $S_2$

EXPLANATION:

$S_1$ in the first 10 seconds:

$\huge{ \boxed{ \bf{ \sf{v = u + at}}}}$

u = 0

a = a m/s²

t = 10 s

v = 0 + a(10)

v = 10a m/s

$\huge{ \boxed{ \bf{ \sf{S = ut + \dfrac{1}{2}at^2}}}}$

$S_1 = 0(t) + \dfrac{1}{2}a(10)^2 \\ \\ \\ S_1 = \frac{100a}{2} \\ \\ \\ S_1 = 50a \: m$

$S_2$ in the last 10 seconds:

After 10s the PARTICLE will have the velocity of v = at which is the final velocity for the first ten seconds and INITIAL velocity for the next ten seconds.

$\huge{ \boxed{ \bf{ \sf{S = ut + \dfrac{1}{2}at^2}}}}$

u = at m/s

a = a m/s²

t = 10 s

$S_2 = at(t) + \dfrac{1}{2}at^2 \\ \\ \\ S_2 =at^2\bigg(1 + \dfrac{1}{2}\bigg) \\ \\ \\ S_2 =a(100)(3/2) \\ \\ \\ S_2 = 150 a \: m\\ \\ \\S_1 = 50a\:\:metres\\ \\ \\S_2 = 150a\:\: metres\\ \\ \\S_2 = 3S_1$

${\large{\bf\textsf{Hence $S_2$ will be three times of $S_1$}}}$