Question

a body moving with constant acceleration travel the distance 3m and 8m respectively in 1s and 2s...Calculate the initial velocity and acceleration of Body​

Answer 1

HEY MATE HERE IS YOUR ANSWER

Using, s=ut+

2

1

at

2

s

1

=u+

2

1

a

or, 3=u+

2

a

∴a=6−2u

s

2

=2u+

2

1

a×4

8=2u+2a

Solving the equations

8=2u+2(6−2u)

8=2u+12−4u

2u=4

u=2m/s

∴a=6−2u=6−4=2m/s

2.

Answer 2

Answer:

♦ The initial velocity of body = 2 m/s

♦ The acceleration of body = 2 m/s²

Step-by-step explanation:

Given:

• Distance travelled, S₁ = 3 m
• Time taken to travel 3 m distance, T₁ = 1s
• Distance travelled, S₂ = 8 m
• Time taken to travel 8 m distance, T₂ = 2s

To find:

• The initial velocity of body.
• The acceleration of body.

Solution:

Let the initial velocity of the body be u.

Using second equation of motion,

★ S = ut + 1/2 at²

Now, in case of S₁ and S₂, we have:

• S₁ = ut₁ + 1/2 at₁². ..1
• S₂ = ut₂ + 1/2 at₂². ..2

Substracting eq 1 from eq 2, we get:

• S₂ - S₁ = ut₁ + 1/2 at₁² - (ut₂ + 1/2 at₂²)
• S₂ - S₁ = u(t₂ - t₁) + 1/2 a(t₂² - t₁²)
• 8 - 3 = u( 2 - 1) + 1/2 a(2² - 1²)
• 8 - 3 = u( 2 - 1) + 1/2 a(4 - 1)
• 5 = u(1) + 1/2 a(3)
• 5 = u + 3/2 a
• a = (10 - 2u)/3 m/s². ...3

Substituting the value of a in equation 1

• 3 = u × 1 + 1/2 × (10 - 2u)/3 × 1
• 3 = u + 1/2 × (10 - 2u)/3 × 1
• 3 = u + (5 - u)/3 × 1
• 3 = (3u + 5 - u)/3 × 1
• 3 = (2u + 5)/3 × 1
• 3 × 3 = (2u + 5)
• 9 = 2u + 5
• 2u = 9 - 5
• 2u = 4
• u = 4/2
• u = 2 m/s

Substituting the value of u in eq3, we have,

• a = (10 - 2 × 2)/3
• a = (10 - 4)/3
• a = 6/3
• a = 2 m/s²

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