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A rollercoaster stops at a point with GPE of 10kJ and then travels down a frictionless slope reaching a speed of 10 m/s at ground level. After this, what length of horizontal track (friction coefficient = 0.5) is needed to bring the rollercoaster to rest?

Recognise that the intial potential energy and kinetic energy at 10 m/s position should be identical due to the frictionless slope. mgh = 0.5mv 2 10kJ = 0.5 x m x 10 2 10 000 = 0.5 x m x 100 50m = 10 000 m =...
DM
Answered by Daniel M. • Maths tutor
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Solve 4cos(2x )+ 2sin(2x) = 1 given -90° < x < 90°. Write 4cos(2x )+ 2sin(2x) in the form Rcos(2x - a), where R and a are constants.

Step 1: Recognise 4cos(2x )+ 2sin(2x) must be put into the form given in the question form i.e. Rcos(2x - a) as it contains only one trigonometric function Step 2: Calculate values of R and a a. Expand Rcos(...
PK
Answered by Prannay K. • Maths tutor
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If cos(x)= 1/3 and x is acute, then find tan(x).

Consider a right angled triangle. Call one of the angles (not the right angle) in this triangle x. We can do this as we are told x is acute. The side opposite to x label O, the side adjacent to x label A, an...
HK
Answered by Hugh K. • Maths tutor
19098 Views

If n is an integer prove (n+3)^(2)-n^(2) is never even.

Let us begin by simplifying the expression:(n+3) 2 - n 2 = (n+3)(n+3) - n 2 = n 2 + 6n + 9 - n 2 (expanded brackets)= 6n + 9 (collected like terms)= 3(2n+3) (taken out a factor of 3)Now we can consider this ...
HK
Answered by Hugh K. • Maths tutor
7581 Views

For what values of k does the graph y=x^(2)+2kx+5 not intersect the x-axis

Where the graph intersects the x-axis, x 2 +2kx+5 must be equal to zero. Thus we can answer the equivalent question: For what k does x 2 +2kx+5 = 0 not have a solution?This is now a simpler problem (roots of...
HK
Answered by Hugh K. • Maths tutor
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