Braking distance - the Australian Mathematical Sciences Institute

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6 ■ maths delivers!The five equations of motionThese equations are for a particle moving in a straight line with a constant acceleration.The history of these equations is not absolutely clear, but we do have some knowledge.In the 14th century, scholastic philosophers at Merton College, Oxford, studied motionwith constant acceleration and deduced what is now known as the Merton rule:An object with constant acceleration travels the same distance as it wouldhave if it had constant velocity equal to the average of its initial and finalvelocities.In the 17th century, Galileo Galilei (1564–1642) and others discovered that, in a void, allfalling objects have the same constant acceleration. Newton (1642–1727) and Leibniz(1646–1716) built on these ideas in developing the concept of the derivative.In the following:• u is the initial velocity of the particle• v is the final velocity of the particle• a is the constant acceleration of the particle• t is the time of motion• x is the distance travelled.The equations are as follows:1 v = u + at2 x = ut + 1 2 at 2(u + v)t3 x =24 v 2 = u 2 + 2ax5 x = vt − 1 2 at 2 .Note. Each of the five equations involve four of the five variables u, v, a, t, x. If the valuesof three of the variables are known, then the remaining values can be found by using twoof the equations.We can use calculus to help derive these five equations.
Braking distance ■ 7The first equation of motionWe assume that ẍ(t) = a, where a is a constant, and that x(0) = 0 and ẋ(0) = u. Integratingẍ(t) = a with respect to t for the first time givesv(t) = ẋ(t) = at + c 1 .Since v(0) = ẋ(0) = u, we have c 1 = u. Hence, we obtain the first equation of motion:v(t) = u + at.The second equation of motionIntegrating again givesx(t) = ut + 1 2 at 2 + c 2 .Since x(0) = 0, we have c 2 = 0. This yields the second equation of motion:x(t) = ut + 1 2 at 2 .The third equation of motionWe start with the second equation of motion:x(t) = ut + 1 2 at 2 22ut + at ut + t(u + at)= = .22The first equation of motion is v = u + at. Substituting, we haveut + vtx(t) =2(u + v)t= .2The fourth equation of motionFrom the first equation, we have t = v − u . Substituting this into the third equation givesa(u + v)tx =2(u + v)(v − u)=2a= v 2 − u 2.2aRearranging to make v 2 the subject produces the fourth equation: v 2 = u 2 + 2ax.
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Page 3 and 4: Introduction . . . . . . . . . . .
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Page 17 and 18: Braking distance ■ 17Exercise 4Fi
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Braking distance - the Australian Mathematical Sciences Institute

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