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The distance of a particle from its start point with respect to time is given by the equation s = 5t^2 + 2t + 4, what is the acceleration of the particle at time t?

To find the velocity from the displacement you must differentiate the function with respect to time.V = (5 2)t^(2-1) + (2 1)t^(1-1) + (4 0)V = 10t +2To find the acceleration you must differentiate the veloci...
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Answered by Louis F. • Maths tutor
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A particle is moving along a straight line. The displacement of the particle from O at time t seconds is s metres where s = 2t^3 – 12t^2 + 7t. Find an expression for the velocity of the particle at time t seconds.

To find the velocity function of a particle when given its displacement function, you must differentiate the given function. v = 6t^2 - 24t + 7 .Similarly, to get the displacement function from the velocity ...
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Answered by Andrew L. • Maths tutor
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Solve the simultaneous equations: x^2 + y^2 = 5 and y = 3x + 1

x 2 + y 2 = 5 1 y = 3x + 1 2 Inserting 2 into 1 : x 2 + (3x +1) 2 = 5 Expanding the brackets: x 2 + 9x 2 + 3x + 3x + 1 = 5Collecting like terms: 10x 2 + 6x - 4 = 0Using the quadratic formula: x = (-6 ± √(6 2...
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Answered by Sophie B. • Maths tutor
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Solve ((3x-2)/4) - ((2x+5)/3) = ((1-x)/6)

The important concept here is that in order to add or subtract fractions, they must have a common denominator. The easiest way of finding a common denominator for three or more fractions is usually to multip...
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Answered by Peta D. • Maths tutor
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Plot the graph, y=2x^2 -7x +4

3 Mehods:Step 1: Does the function factorise. In this case yes it does. giving y=(2x+1)(x-4)Method: Find the factors of the contract in this case (4). Which are (2,2) , (4,1)We know that a quadratic must fac...
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Answered by Theodore W. • Maths tutor
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