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A bird is flying with an initial velocity of 10 m/s. It accelerates uniformly and covers a distance of 25 m to reach a final velocity of 15 m/s. Calculate the acceleration of the bird.

Physics

ScanSolve AI Answer

Answer

c) 2.5 m/s2\text{c) } 2.5 \text{ m/s}^2

Step-by-step solution

5 steps
  1. Identify the given values.

    Initial velocity, u=10 m/su = 10 \text{ m/s}
    Final velocity, v=15 m/sv = 15 \text{ m/s}
    Distance covered, s=25 ms = 25 \text{ m}

  2. Choose the appropriate kinematic equation.

    We need to find the acceleration (aa). The kinematic equation that relates initial velocity, final velocity, acceleration, and distance is:

    v2=u2+2asv^2 = u^2 + 2as
  3. Substitute the values into the equation.

    (15 m/s)2=(10 m/s)2+2×a×(25 m)(15 \text{ m/s})^2 = (10 \text{ m/s})^2 + 2 \times a \times (25 \text{ m}) 225 m2/s2=100 m2/s2+50a m225 \text{ m}^2/\text{s}^2 = 100 \text{ m}^2/\text{s}^2 + 50a \text{ m}
  4. Solve for acceleration (aa).

    Subtract 100 m2/s2100 \text{ m}^2/\text{s}^2 from both sides:

    225 m2/s2−100 m2/s2=50a m225 \text{ m}^2/\text{s}^2 - 100 \text{ m}^2/\text{s}^2 = 50a \text{ m} 125 m2/s2=50a m125 \text{ m}^2/\text{s}^2 = 50a \text{ m}

    Divide by 50 m50 \text{ m}:

    a=125 m2/s250 ma = \frac{125 \text{ m}^2/\text{s}^2}{50 \text{ m}} a=12550 m/s2a = \frac{125}{50} \text{ m/s}^2 a=2.5 m/s2a = 2.5 \text{ m/s}^2
  5. Compare with the given options.

    The calculated acceleration is 2.5 m/s22.5 \text{ m/s}^2, which corresponds to option (c).

    The magnitude of the acceleration of the bird is c) 2.5 m/s2\text{c) } 2.5 \text{ m/s}^2.

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