Question

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A body travels 200 cm in the first 2 s and 220 cm in the next 5 s. Calculate the velocity at the end of seventh second from the start. (in m\s)...

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Answer 1

Answer :-

◾As it's Given

▪️Body travels 200 cm in first 2 sec

▪️And 220 cm in next 5 sec

◾So from above we came to know

⏩Body travels 420 cm in 7 sec

◾Now we will write equation for both case

⏩S = ut + 1/2 at²

▪️For 200 cm (2 m)

→2 = u(2) + 1/2 × a × 2²

→ 2 = 2u + 2a

→1 = u + a .......(i)

▪️For 420 cm (4.2 m)

→4.2 = u(7) + 1/2 × a × 7²

→4.2 = 7u + 49/2 × a

or

→0.6 = u + 7/2 × a

→ 0.6 = u + 3.5a .....(ii)

◾Now (i) - (ii)

1 = u + a

- 0.6 = u + 3.5a

___________________

0.4 = - 2.5a

Or

$\implies a = \dfrac{0.4}{-2.5}$

$\implies a = \dfrac{-16}{100} \: ms^{-2}$

= - 0.16 m/s²

◾Now by putting value of a in (i)

$\implies 1 = u + (- 0.16)$

$\implies 1 = u - 0.16$

$\implies u = 1 + 0.16$

$\implies u = 1.16$

So u = 1.16 m/s

◾Now as

⏩v = u + at

◾Where

▪️$u = 1.16\: m/s$

▪️$t = 7 \: sec$

▪️$a = -0.16 \: m/s^2$

◾Then velocity at end of 7th sec

$\implies v = 1.16 + 7 \times(- 0.16)$

$\implies v = 1.16 - 1.12$

$\implies v = 0.04 m/s$

◾So velocity at end of 7th sec = 0.04 m/s

Answer 2

Velocity first =0 and 200cm in 2 seconds

Use this formula

{V1=u+a1.t=0+a 1.2}

hence, it is V1-2.a1

It covered 200cm know just 2 seconds

200=u.t+(1/2).a1.t^2=0+(1/2 )a1.4=2.a

hence, there a1=200/2=100/S^2

In just 5 seconds it covered 220 cm

a2=5=V1.t+(1/2).a2.(t^2)

200=200.5+(1/2).a2.(5^2)

here here now,

A2=2 and (220-1000)/25

so,

v2 = v1 + a2. 5 = 200 + { -312/5 }.5

$answer \: = - 122m \: per \: sec$