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A body starts from rest and accelerates at 2 m/s² and then retards at 3 m/s² to come to rest. If the total time taken is 10 s, find the maximum speed attained.

Physics

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Answer

12 m/s\text{12 m/s}

Step-by-step solution

4 steps

  1. Identify the given values.

    The body starts from rest and accelerates with a1=2 m/s2a_1 = 2 \text{ m/s}^2.
    It then retards to rest with a2=3 m/s2a_2 = 3 \text{ m/s}^2.
    The total time taken is t=10 st = 10 \text{ s}.

  2. Use the formula for maximum velocity.

    When a body starts from rest, accelerates with a1a_1 for time t1t_1 to reach maximum velocity vmaxv_{max}, and then retards with a2a_2 for time t2t_2 to come to rest, the maximum velocity attained is given by the formula:

    vmax=a1a2a1+a2tv_{max} = \frac{a_1 a_2}{a_1 + a_2} t

  3. Substitute the given values into the formula.

    vmax=2 m/s2×3 m/s22 m/s2+3 m/s2×10 sv_{max} = \frac{2 \text{ m/s}^2 \times 3 \text{ m/s}^2}{2 \text{ m/s}^2 + 3 \text{ m/s}^2} \times 10 \text{ s}

  4. Calculate the maximum speed.

    vmax=6 m2/s45 m/s2×10 sv_{max} = \frac{6 \text{ m}^2/\text{s}^4}{5 \text{ m/s}^2} \times 10 \text{ s} vmax=65×10 m/sv_{max} = \frac{6}{5} \times 10 \text{ m/s} vmax=1.2×10 m/sv_{max} = 1.2 \times 10 \text{ m/s} vmax=12 m/sv_{max} = 12 \text{ m/s}

    The maximum speed attained is 12 m/s12 \text{ m/s}. This corresponds to option (1).

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