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The displacement h metres of a particle moving along a straight line after t seconds is given by h = -2t³ + 3/2 t² + 3t. (a) Find the initial acceleration.

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3 parts
a

3 m/s23 \text{ m/s}^2

b(i)

1 s1 \text{ s}

b(ii)

2.5 m2.5 \text{ m}

278 m/s\frac{27}{8} \text{ m/s}

Step-by-step solution

6 steps

The displacement of the particle is given by the function:

h=−2t3+32t2+3th = -2t^3 + \frac{3}{2}t^2 + 3t

Step 1: Find the velocity function.
Velocity vv is the first derivative of displacement hh with respect to time tt:

v=dhdt=ddt(−2t3+32t2+3t)v = \frac{dh}{dt} = \frac{d}{dt}\left(-2t^3 + \frac{3}{2}t^2 + 3t\right) v=−6t2+3t+3v = -6t^2 + 3t + 3

Step 2: Find the acceleration function.
Acceleration aa is the first derivative of velocity vv with respect to time tt:

a=dvdt=ddt(−6t2+3t+3)a = \frac{dv}{dt} = \frac{d}{dt}(-6t^2 + 3t + 3) a=−12t+3a = -12t + 3

Part a· 1 step

Find the initial acceleration.

  1. Calculate the acceleration at t=0t=0.

    Initial acceleration occurs at t=0t=0. Substitute t=0t=0 into the acceleration function:

    a(0)=−12(0)+3a(0) = -12(0) + 3 a(0)=3 m/s2a(0) = 3 \text{ m/s}^2

    The initial acceleration is 3 m/s2\boxed{3 \text{ m/s}^2}.

Part b(i)

1 s1 \text{ s}

The time when the particle was momentarily at rest.

Calculate

  1. Set velocity to zero and solve for tt.

    The particle is momentarily at rest when its velocity vv is 0:

    −6t2+3t+3=0-6t^2 + 3t + 3 = 0

    Divide the equation by −3-3:

    2t2−t−1=02t^2 - t - 1 = 0

    Factor the quadratic equation:

    (2t+1)(t−1)=0(2t + 1)(t - 1) = 0

    This gives two possible values for tt:

    2t+1=0  ⟹  t=−122t + 1 = 0 \implies t = -\frac{1}{2} t−1=0  ⟹  t=1t - 1 = 0 \implies t = 1

    Since time tt cannot be negative in this context, we take the positive value.
    The time when the particle was momentarily at rest is 1 s\boxed{1 \text{ s}}.

Part b(ii)

2.5 m2.5 \text{ m}

Its displacement by the time it comes to rest momentarily.

  1. Calculate displacement at t=1t=1.

    Substitute t=1t=1 second (from part b(i)) into the original displacement function:

    h(1)=−2(1)3+32(1)2+3(1)h(1) = -2(1)^3 + \frac{3}{2}(1)^2 + 3(1) h(1)=−2+32+3h(1) = -2 + \frac{3}{2} + 3 h(1)=1+32h(1) = 1 + \frac{3}{2} h(1)=52 mh(1) = \frac{5}{2} \text{ m}

    The displacement by the time it comes to rest momentarily is 2.5 m\boxed{2.5 \text{ m}}.

    Calculate the maximum speed attained.

  2. Find the time when acceleration is zero.

    The maximum positive velocity occurs when the acceleration is zero. Set the acceleration function to 0:

    a=−12t+3=0a = -12t + 3 = 0 −12t=−3-12t = -3 t=14 st = \frac{1}{4} \text{ s}

  3. Calculate the velocity at this time.

    Substitute t=14t = \frac{1}{4} into the velocity function:

    v(14)=−6(14)2+3(14)+3v\left(\frac{1}{4}\right) = -6\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) + 3 v(14)=−6(116)+34+3v\left(\frac{1}{4}\right) = -6\left(\frac{1}{16}\right) + \frac{3}{4} + 3 v(14)=−38+68+248v\left(\frac{1}{4}\right) = -\frac{3}{8} + \frac{6}{8} + \frac{24}{8} v(14)=−3+6+248=278 m/sv\left(\frac{1}{4}\right) = \frac{-3 + 6 + 24}{8} = \frac{27}{8} \text{ m/s}

  4. Determine the maximum speed.

    The velocity function v(t)=−6t2+3t+3v(t) = -6t^2 + 3t + 3 is a downward-opening parabola. Its maximum value for t≥0t \ge 0 occurs at its vertex, t=14t = \frac{1}{4} s. At this point, the velocity is positive. For t>1t > 1 s, the velocity becomes negative and its magnitude increases indefinitely. Therefore, the "maximum speed attained" refers to the maximum positive velocity.
    The maximum speed attained is 278 m/s\boxed{\frac{27}{8} \text{ m/s}}.

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