# Tutor profile: Neil L.

## Questions

### Subject: Physics

Using the gravitational force equation and the circular motion equations, show Kepler's 3rd Law ( Kepler's 3rd law is T^2 = KR^3 where R = distance from planet to star, and T is the orbital period.) to be true. Also find an expression for K.

Gravitational force equation: F = GMm/(R^2) Where G is the gravitational constant, M is the mass of the star and m is the mass of the planet. Circular Motion equation: F = mRω^2 Therefore, equating equations: mRω^2 = GMm / (R^2) (Cancelling m) Rω^2 = GM / (R^2) (Substituting ω = 2π / T) 4Rπ^2 / (T^2) = GM / R(^2) Therefore, rearranging: T^2 = (R^3) (4π^2) / GM Therefore K = (4π^2) / GM

### Subject: Physics

A truck travels along a straight, rough road. The truck pulls a trailer with a mass of 500kg, and the truck has a mass of 1000kg. The truck is joined to the trailer by a light, inextensible rod. The truck feels a frictional force from the road of 500N, and the trailer feels a frictional force of 300N. a) What driving force is required to maintain a constant velocity? Assume there is a net force of 200N in the positive direction b) Find the acceleration of the truck-trailer system c) Find the tensile force (tension) in the rod joining the truck and trailer d) What must the driving force exerted by the truck be?

a) Newton's Second Law: F = ma Acceleration must be equal to 0 for constant velocity Therefore consider the total mass and the total frictional force: The total mass is 1000+500 = 1500kg Total frictonal force = 800N Friction always opposes the motion Driving force must equal frictional force, therefore F = 800N b) Consider again Newton's Second Law: F = ma Therefore 200 = 1500a So a = 200/1500 = 2/15 (or decimal form) c) Consider only the trailer. The frictional force on the trailer is 300N, and the net force is 200N. The driving force is delivered through the rod, so to find the tension, we take the direction of the rod's tension, T, as positive: T - 300 = 200, T = 500 d) Consider now the truck: The net force on the truck is also 200N. Tension is equal on both sides of the rod, and since the truck is pulling the trailer ( and so is in front of it ), the tension is in the same direction as the frictional force. As such, the total force in the direction of friction is 500 + 500 = 1000N Now consider the net force of 200N again. This means the driving force, D, is given by: D - 1000 = 200 D = 1200

### Subject: Calculus

Given the function f(x) = ( x^2 ) ( ( 4x^3 + 7 ) ^-1 ), find: a) ∫f(x) dx b) f'(x) Keeping both in their simplest forms and explaining your method.

a) Using the standard result: a ∫( f(x)/f'(x) ) dx = ( 1/a ) ( ln ( f'(x) ) Therefore, let ( x^2 ) = b, and let ( 4x^3 + 7) = c, db/dx = 2x, dc/dx = 12x^2 Hence we can see the denominator, c, is a multiple of the numerator, b. Thus, from the standard result, we can see that: a = 1/12, and that ∫f(x) = aln(c) Hence ∫f(x) = 12ln(4x^3 + 7) b) Using the quotient rule (product rule is also possible but takes much longer) Quotient rule: f(x) = g(x)/h(x) --> f'(x) = (g'(x) h(x) - h'(x) g(x)) / ( h(x) )^2 Therefore, using the differentials found for part (a), and letting g(x) = b and h(x) = c, we know: g'(x) = 2x, h'(x) = 12x^2 Therefore substituting these results into the quotient rule formula gives: f'(x) = ( 2x(4x^3 + 7) - (12x^2)(x^2) ) / ((4x^3 + 7)^2) Expansion of the denominator does not allow for any further simplification, as such the answer can either contain the expanded or factorised form of the following final answer: f'(x) = (14x - 4x^4) / ((4x^3 + 7)^2)

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