Question

While moving with uniform acceleration, a body has covered 100m in 10th second and
attained a velocity of 105 m/s. Its initial velocity u' and acceleration a' respectively are
1) 10 ms-1 , 5 m-2
210 ms-1 ,5 ms-2
3) 5 ms-1 , 10 ms-2
4) 10 ms-1 , 0 ms-2​

Answer 1

Answer:

v=u+at

105=u+10a .....(1)

550=u(10)+

2

1

×a×10

2

⇒550=10u+50a⇒55=u+5a .....(2)

eq

n

(1)−eq

n

(2)

⇒50=5a

a=10 m/s

2

u=5 m/s

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Answer 2

s = 100 m

t = 10secs

n = 10

v = 105m/s

By using First equation of motion

v = u + at

105 = u + a(10)

105 = u + 10a - eq (1)

By using Displacement in 'n^th' Second of Motion

s = u + a/2 (2n-1)

100 = u + a/2 (2(10)-1)

100 = u + a/2(19)

100 = u + 19a/2 - eq (2)

By equating eq - (1) & (2)

105 = u + 10a rough work

100 = u + 19a/2 = 10a-19a/2

- - - = 20a/2-19a/2

5 = 1/2 a =1/2a

5×2 = a

a = 10m/s^2

Put the value of a = 10m/s^2 in eq - (1)

105 = u + 10a

105 = u + 10(10)

105 = u + 100

105 - 100 = u

u = 5m/s

So, a = 10m/s^2 & u = 5m/s

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