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Use the logarithm tables to evaluate If T = 2 √G π , make G the subject of the formula. Hence, find the value of G to the nearest whole number when T=45, L=8 and π = 3.142.

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Answers

4 parts
a

3\text{3}

b(i)

208 units/s\text{208 units/s}

b(ii)

112 units/s2\text{112 units/s}^2

c

12\text{12}

Step-by-step solution

8 steps

Part a· 2 steps

To find the value of P, we first find the derivative of the function yy with respect to xx. A stationary value occurs when the first derivative is equal to zero.

  1. Differentiate y=2x4−Px2+2xy = 2x^4 - Px^2 + 2x with respect to xx.

    dydx=ddx(2x4−Px2+2x)=8x3−2Px+2\frac{dy}{dx} = \frac{d}{dx}(2x^4 - Px^2 + 2x) = 8x^3 - 2Px + 2

  2. Set dydx=0\frac{dy}{dx} = 0 and substitute x=−1x = -1 to solve for P.

    8(−1)3−2P(−1)+2=08(-1)^3 - 2P(-1) + 2 = 0 −8+2P+2=0-8 + 2P + 2 = 0 2P−6=02P - 6 = 0 2P=62P = 6 P=3P = 3

    The value of P is 3\boxed{\text{3}}.

Part b(i)

208 units/s\text{208 units/s}

Velocity

The position of the particle is given by x=5t3−4t2x = 5t^3 - 4t^2.

  1. Find the velocity function by differentiating the position function with respect to time.

    v(t)=dxdt=ddt(5t3−4t2)=15t2−8tv(t) = \frac{dx}{dt} = \frac{d}{dt}(5t^3 - 4t^2) = 15t^2 - 8t

  2. Calculate the velocity at t=4t = 4 seconds.

    v(4)=15(4)2−8(4)v(4) = 15(4)^2 - 8(4) v(4)=15(16)−32v(4) = 15(16) - 32 v(4)=240−32v(4) = 240 - 32 v(4)=208v(4) = 208

    The velocity after 4 seconds is 208 units/s\boxed{\text{208 units/s}}.

Part b(ii)

112 units/s2\text{112 units/s}^2

Acceleration

  1. Find the acceleration function by differentiating the velocity function with respect to time.

    a(t)=dvdt=ddt(15t2−8t)=30t−8a(t) = \frac{dv}{dt} = \frac{d}{dt}(15t^2 - 8t) = 30t - 8

  2. Calculate the acceleration at t=4t = 4 seconds.

    a(4)=30(4)−8a(4) = 30(4) - 8 a(4)=120−8a(4) = 120 - 8 a(4)=112a(4) = 112

    The acceleration after 4 seconds is 112 units/s2\boxed{\text{112 units/s}^2}.

Part c

12\text{12}

To evaluate the definite integral, we first find the antiderivative of the function and then apply the limits of integration.

  1. Find the indefinite integral of 6x2−4x+26x^2 - 4x + 2.

    ∫(6x2−4x+2) dx=6x2+12+1−4x1+11+1+2x0+10+1+C\int (6x^2 - 4x + 2) \,dx = \frac{6x^{2+1}}{2+1} - \frac{4x^{1+1}}{1+1} + \frac{2x^{0+1}}{0+1} + C =6x33−4x22+2x+C= \frac{6x^3}{3} - \frac{4x^2}{2} + 2x + C =2x3−2x2+2x+C= 2x^3 - 2x^2 + 2x + C

  2. Evaluate the definite integral from 0 to 2.

    ∫02(6x2−4x+2) dx=[2x3−2x2+2x]02\int_{0}^{2} (6x^2 - 4x + 2) \,dx = \left[ 2x^3 - 2x^2 + 2x \right]_{0}^{2} =(2(2)3−2(2)2+2(2))−(2(0)3−2(0)2+2(0))= (2(2)^3 - 2(2)^2 + 2(2)) - (2(0)^3 - 2(0)^2 + 2(0)) =(2(8)−2(4)+4)−(0−0+0)= (2(8) - 2(4) + 4) - (0 - 0 + 0) =(16−8+4)−0= (16 - 8 + 4) - 0 =8+4= 8 + 4 =12= 12

    The value of the integral is 12\boxed{\text{12}}.

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