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A body moves, from rest with a constant acceleration of 5 m per sec. The distance covered in 5 sec is most nearly

A. 38 m

B. 62.5 m

C. 96 m

D. 124 m

Answer: Option B


This Question Belongs to Civil Engineering >> Strength Of Materials

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Comments ( 5 )

  1. Umati Umati
    Umati Umati :
    3 years ago

    @infinite knowledge sol#
    Body start from rest means "Initial velocity is zero".The body start moving with acceleration 5 meter/sec.how much distance covered by body in 5 sec ?
    Initial velocity is represnted by " vi"=0
    time is "t". T=5
    acceleration is "a"=5
    distance is "s".
    so, formula for distance according to question is
    S=Vi×t+1/2at^2
    S=0×5+1/2×5×5^2
    S=0+1/2×5×25
    S=0+1/2×125
    S=0+0.5×125
    S=62.5 and.

  2. SANEET KASHYAP
    SANEET KASHYAP :
    4 years ago

    Body starts from rest u=0
    time = 5sec
    acc = 5m/s^2
    s=ut+1/2at^2
    s= 0+(5*(5*5))/2=125/2=62.5m

  3. Manhar Bharvad
    Manhar Bharvad :
    4 years ago

    Please explain how?

  4. Study Mithra
    Study Mithra :
    5 years ago

    @infinite knowledge use below formula

    s=ut+1/2at^2

    where u =0 m/s
    t = 5 sec
    a = 5m/s^2

  5. Infinite Knowledge
    Infinite Knowledge :
    5 years ago

    Why the answer is B

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