Physics Help!!!!?

A car enters the freeway with a speed of 7.1 m/s and accelerates uniformly for 2.6 km in 3.8 min.

How fast is the car moving after this time? Answer in units of m/s.

3 Answers

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  • 2 months ago
    Best Answer

    This is just a straightforward kinematics problem.  You just need to take care with the units.

     I use the suvat symbols. s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time.  You might use different symbols, but you should recognise the form of the equations.

    s = ut + ½at²

    v = u + at

    v² = u² + 2as

    s = ½(u +v) * t

    s = 2.6 km = 2600 m

    u = 7.1 m/s

    v = ?

    a = {not given}

    t = 3.8 m = (3.8 * 60) = 228 seconds

    s = ½(u +v) * t

    Plug in values:

    2600 = ½(7.1 +v) * 228

    v = 15.7 m/s

    {16 m/s to two significant figures.}

  • oubaas
    Lv 7
    2 months ago

    t = 3.8*60 = 228 sec

    Vm = d/t = 2600/228 = 11.405 m/sec

    2Vm = Vi+Vf

    Vf = 2Vm-Vi = 22.81-7.1 = 15.71 m/sec

  • 2 months ago

    Lets get some consistent units for a start.

    He moves 2600 m in a time of 3.8 * 60 = 228 seconds

    the final speed = 7.1 + at

    the average speed = (U+v) / 2 = ( 7.1 +(7.1+ a * 228) ) /2

    = 7.1 + a * 228 / 2 = 7.1+114 a

    Now as speed = distance / time

    7.1 + 114 a = 2600 / 228

    find a

    subtract 7.1 from both sides

    114 a = ( 2600/228) - 7.1

    divide both sides by 114

    a = ( ( 2600/228) - 7.1) / 114 = 0.038 m/s^2

    There are many other mathematical approaches that yield the same result.

    As a check note that the average speed needed is about 11.4 m/s

    at the acceleration I calculated the car GAINS about 8.6 m/s

    so it started at 7.1 and finished at 15.7 which gives you the average of about 11.4

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