Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

Given velocity function, find acceleration and distance?

The speed of a car (km/h) is given by the expression

v(t) = 2t^2+ 5t, 0 < t < 1

where t is time in hours. Use this expression to find

(a) The acceleration over this time period.

(b) The total displacement over the same time period

4 Answers

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  • 9 years ago
    Favorite Answer

    Hi Hfgf,

    -----

    (a)

    To get acceleration from velocity, you need to differentiate with respect to time, because acceleration describes how velocity changes over time.

    a(t) = d/dt[v(t)] = d/dt[2t² + 5t] = d/dt[2t²] + d/dt[5t] = 2 d/dt[t²] + 5 d/dt[t]= 2(2t) + 5(1) = 4t + 5.

    This is a linear function, so the graph of acceleration will be a straight line.

    At t = 0:

    a(0) = 4(0) + 5 = 0 + 5 = 5 km/h².

    a(1) = 4(1) + 5 = 4 + 5 = 9 km/h².

    So, you can draw the acceleration over this time period. It will be a straight line with gradient 4:

    http://www3.wolframalpha.com/input/?i=plot+4t%2B5+...

    -----

    (b)

    To get displacement from velocity, you need to integrate with respect to time, because velocity describes how displacement changes over time.

    s(t) = ∫ v(t) dt = ∫ (2t² + 5t) dt = ∫ 2t² dt + ∫ 5t dt = 2 ∫ t² dt + 5 ∫ t dt = 2 (t³ / 3) + 5(t² / 2) + constant.

    = (2t³ / 3) + (5t² / 2) + constant.

    To find the total displacement over the time period, use the limits of integration from 0 to 1:

    [(2(1³) / 3) + (5(1²) / 2)] - [(2(0³) / 3) + (5(0²) / 2)]

    = [(2 / 3) + (5 / 2)] - [(0 / 3) + (0 / 2)]

    = [(4 / 6) + (15 / 6)] - [0 + 0]

    = [19 / 6] - [0]

    = 19 / 6

    = 3.1666666... km.

    Here is the graph of the displacement over that time:

    http://www3.wolframalpha.com/input/?i=plot+%282t%C...

  • dennis
    Lv 6
    9 years ago

    a(t) = d/dt( v(t) = 4t +5

    Displacement = Integral v(t) dt between limits of 0 and 1

    = [ 2t^3/3 +5t^2/2] between 0 and 1= 2/3 +5/2 = 19/6 km.

  • TomV
    Lv 7
    9 years ago

    a) a(t) = dv/dt = 4t+5

    b) s(t) = integral(2t²+5t) = 2t³/3 + 5t²/2 + C

    Assuming the initial condition s(0) = 0, C = 0

    s(1) = 2/3 + 5/2 = 19/6

  • 9 years ago

    (a)

    a(t) = v'(t) = 4t + 5

    If 0 < t < 1, then a(t) will increase from 5 to 9 during that period.

    (b)

    Integrate v(t) to get the displacement s(t):

    s(t) = (2/3)t^3 + (5/2)t^2 + C

    So we want:

    s(1) - s(0)

    = ((2/3)(1^3) + (5/2)(1^2) + C) - ((2/3)(0^3) + (5/2)(0^2) + C)

    = (2/3)(1) + (5/2)(1) + C - (2/3)(0) - (5/2)(0) - C

    = (2/3) + (5/2) - 0 - 0

    = (4/6) + (15/6)

    = (4 + 15) / 6

    = 19/6

    =~ 3.1666666666666...

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