a jeep start from the state of rest .if ots velocity became 60km/hr in 15 min. (i) what is the acceleration of the jeep?(ii) what is the

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a jeep start from the state of rest .if ots velocity became 60km/hr in 15 min. (i) what is the acceleration of the jeep?(ii) what is the distance coverd by jeep?​

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3 months 2021-07-23T17:29:28+00:00 1 Answers 45 views 0

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    0
    2021-07-23T17:30:58+00:00

    Explanation:

    Here,

    • Initial velocity (u) = 0 (as it starts from rest)
    • Final velocity (v) = 60 km/h
    • Time taken (t) = 15 min

    Converting the quantities into its standard form :

    → v = 60 km/h

    • 1 km/h = 5/18 m/s

    → v = ( 60 ×  \sf \dfrac{5}{18} ) m/s

    → v = ( 10 ×  \sf \dfrac{5}{3} ) m/s

    → v =  \sf \dfrac{50}{3} m/s

    v = 16.66 m/s

    Also,

    → t = 15 minutes

    → t = (15 × 60) seconds

    t = 900 seconds

    Calculating acceleration :

    v = u + at

    • v is final velocity
    • a is acceleration
    • u is initial velocity
    • t is time

    → 16.66 = 0 + 900a

    → 16.66 – 0 = 900a

    → 16.66 = 900a

     \sf \dfrac{16.66}{900} = a

    0.0185 m/s² = a

    Acceleration is 0.0185 m/s².

    _________________________

    Calculating distance covered :

    v² u² = 2as

    • v is final velocity
    • a is acceleration
    • u is initial velocity
    • s is distance

    → (16.66)² – (0)² = 2 × 0.0185 × s

    → 277.5556 – 0 = 0.037s

    → 277.5556 = 0.037s

     \sf \dfrac{277.5556}{0.037} = s

    7501.50 m = s

    Distance covered is 7501.50 m .

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