A snowmobile has an initial velocity of +4.4 m/s.If it accelerates at the rate of +0.34 m/s2 for 4.2 s, what is the final velocity?

Answer in units of m/s.

If instead it accelerates at the rate of −0.78 m/s2, how long will it take to reach a complete stop?Answer in units of s.

4 Answers

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  • NCS
    Lv 7
    1 month ago
    Favorite Answer

    Use v = v₀ + a*t

    Given v₀ = 4.4 m/s

    then

    a) when t = 4.2 s and a =0.34 m/s²,

    v = 4.4m/s + 0.34m/s² * 4.2s = 5.8 m/s

    b) and when a = -0.78 m/s² and v = 0 ("reach a ... stop")

    0 = 4.4m/s - 0.78m/s² * t

    t = 5.6 s

    Both answers to two significant digits, matching the data.

    Hope this helps!

  • 1 month ago

    We can use v = u + at

    so, v = 4.4 + 0.34(4.2) => 5.8 m/s

    Again, using v = u + at we have,

    0 = 4.4 + (-0.78)t

    so, t = 4.4/0.78 => 5.6 seconds

    :)>

  • oubaas
    Lv 7
    1 month ago

    Vf = Vi+a*t = 4.4+0.34*4.2 = 5.8 m/sec 

    ts = (0-Vi)/a = -4.4/-0.78 = 5.6 sec 

    all answers with just 2 significant digits

  • Jim
    Lv 7
    1 month ago

    v = at + v₀

    v = .34(4.2) + 4.4 = 5.828m/s (rounds to 5.8m/s)

    Deceleration, using same formula:

    0 = (-0.78)t +4.4 ((assuming 4.4 initial velocity))

    t = 5.64103s (but rounds to 5.6s)

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