Car at 79km/hr hits brakes and comes to stop in 11.3 seconds. What was car’s deceleration and how far did it go before it came to a stop?

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  • NCS
    Lv 7
    1 month ago
    Favorite Answer

    79km/h = 21.94 m/s

    acceleration a = -21.94m/s / 11.3s = -1.94 m/s²

    and for "deceleration" you probably don't need the sign.

    distance s = Vavg * t = ½ * 21.94m/s * 11.3s = 124 m

    or

    s = v0+t + ½at²

    or

    v² = 0 = u² + 2as

    all give the same result

    If you find this helpful, please award Best Answer!

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  • MyRank
    Lv 6
    3 weeks ago

    Given,

    Time (t) = 11.3sec

    Deceleration (a) = ?

    Distance (d) = ?

    Final speed (u) = 0m/sec

    Initial speed (v) = 79km/hr = 21.94m/sec

    Using kinematic relation:-

    Acceleration (a) = v-u/t

    = (0-21.94)m/sec / 11.3sec = -21.94/11.3 m/sec²

    = -1.94m/sec²

    ∴ Acceleration (a) = -1.94m/sec²

    Now, v²= u²+2ad

    (0m/sec)² = (21.94m/sec)² + 2(-1.94m/sec²) x d

    = (21.94)²m²/sec² = 3.88m/sec² x d

    = 481.3636m²/sec² = 3.88m/sec² x d

    = Distance (d) = 481.3636/3.88 = 124m

    ∴ Distance (d) = 124m.

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  • Anonymous
    1 month ago

    Unanswerable regarding travel distance.

    Even Courts & Police have a difficulty in establishing "skid" distance re speed.

    Tyre Friction, and road friction coefficient is nearly a science unto itself.

    Damp pavement, Icy pavement, concrete, MacCadam,  gravel, tar,  leaves, sand, etc.

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  • Anonymous
    1 month ago

    acceleration = change in speed / time = 6.99 km / hr-sec

    distance = 1/2 * acceleration * time^2

    • jennifer1 month agoReport

      How’s would you find the displacement?

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