Question

5. A particle covers 10m in first 5s and and 10m in next 3s. Assuming constant acceleration, find initial

speed, acceleration and distance covered in next 2s.

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Answer 1

Answer:

we know that initial velocity of a particle will be 0.

1)given time =5 seconds, Distance =10 m/s

s=ut+1/2a*t^2

10=u×5+1/2*a*(5)^2

10=5u+25a/2

10-(5u+25a/2)=0

-5(2u+5a-4)=0

2u+5a-4=0

2u +5a=4.......(1)

(2) given time =5+3=8 seconds, Distance =20m

s=ut+1/2a*t^2

20=8u+1/2*a(8)^2

20=8u+64a/2

20-(8u+64a/2)=0

20-8u-32a=0

-4(2u+8a-5)=0

2u+8a-5=0

2u+8a=5........(2)

on solving (1) and(2) we get

4-5=2u+5a-2u-8a

-1=-3a

a=1/3.....(3)

substitute (3) in (2) we get

2u+8a=5

2u+8(1/3)=5

2u+8/3=5

6u+8=15

6u=7

u=7/6.........(4)

therefore the distance covered in first 10 seconds

s=ut+1/2at^2

=7/6*10+1/2*1/3(10)^2

=70/6+100/6

=170/6

=28.3m

therefore the distance traveled by the particle in next 2 seconds

=28.3-20

=8.3m

Explanation:

please do follow me

Answer 2

Given :-

In 1st case,

☄ initial Velocity, u = 0

☄ distance covered, s = 10

☄ time taken, t = 5s

In 2nd case,

☄ initial Velocity, u = 0

☄ distance covered, s = 10

☄ time taken, t = 3s

To find :-

speed, accerlation and distance covered by the particle

Solution :-

using equation of motion .i.e.,

In 1st case

${\dashrightarrow{ \bf s=ut+ \dfrac{1}{2} a {t}^{2}}}$

${\dashrightarrow {\sf 10=(u)(5)+ \dfrac{1}{2}( a)(5) {}^{2} }}$

${\dashrightarrow {\sf 10=5u+\dfrac{25a}{2} }}$

${\dashrightarrow {\sf 10- \bigg(5u+ \dfrac{25a}{2} \bigg)=0}}$

${\dashrightarrow {\sf -5(2u+5a-4)=0}}$

${\dashrightarrow {\sf 2u+5a-4=0}}$

${\dashrightarrow {\sf 2u +5a=4}} \: \: \: \: .......(1)$

In 2nd case

${\dashrightarrow {\bf s=ut+ \dfrac{1}{2} at^2}}$

${\dashrightarrow {\sf 20=8u+ \dfrac{1}{2} a(8)^2}}$

${\dashrightarrow {\sf 20=8u +\dfrac{64a}{2}}}$

${ \dashrightarrow {\sf 20- \bigg(8u+ \dfrac{64a}{2} \bigg)=0}}$

${\dashrightarrow {\sf 20-8u-32a=0}}$

${\dashrightarrow {\sf -4(2u+8a-5)=0}}$

${ \dashrightarrow {\sf 2u+8a-5=0}}$

${ \dashrightarrow {\sf 2u+8a=5}} \: \: \: \: .....(2)$

Solving eq (1) and (2) we get

$\leadsto \sf 4-5= \cancel{2u}+5a \: \cancel{-2u} \: -8a$

$\leadsto \sf -1=-3a$

$\leadsto \sf a= \dfrac{1}{3}$

thus, accerlation of the particle is 1/3 m/s²

Now,

put a=1/3 in eq (2)

$\leadsto \sf 2u+8a=5$

$\leadsto \sf 2u+(8) \dfrac{1}{3} =5$

$\leadsto \sf 2u+ \dfrac{8}{3} =5$

$\leadsto \sf 6u+8=15$

$\leadsto \sf 6u=7$

[tex\leadsto \sf u= \dfrac{7}{6} [/tex]

the distance covered in first 10 second

$\dashrightarrow \sf s=ut+ \dfrac{1}{2} at^2$

$\dashrightarrow \sf s= \dfrac{7}{6} \times 10+ \dfrac{1}{2} \times \dfrac{1}{3} (10)^2$

$\dashrightarrow \sf s= \dfrac{70}{6} + \dfrac{100}{6}$

$\dashrightarrow \sf s= \dfrac{170}{6}$

$\dashrightarrow \sf s=28.3m$

hence, the distance traveled by the particle in next 2s =28.3 - 20 = 8.3m

$\:$

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