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Physics 114 – Lecture 4

Physics 114 – Lecture 4. 2.5 Motion with Constant Acceleration Reminder: v ave = Δ x/ Δ t = (x 2 – x 1 )/(t 2 - t 1 ) and a ave = Δ v/ Δ t = (v 2 – v 1 )/(t 2 - t 1 ) We now write the final velocity as v, that is v 2 → v and the final time as t, that is t 2 → t.

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Physics 114 – Lecture 4

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  1. Physics 114 – Lecture 4 2.5 Motion with Constant Acceleration Reminder: vave = Δx/ Δt = (x2 – x1)/(t2 - t1) and aave = Δv/ Δt = (v2 – v1)/(t2 - t1) We now write the final velocity as v, that is v2 → v and the final time as t, that is t2 → t. Likewise, we now write the initial velocity as v0, that is v1 → v0 , and the initial time as t0 , that is t1 → t0 , and treat the final and initial values of x in the same way, that is x2 → x and x1 → x0 . L4-s1,12

  2. Physics 114 – Lecture 4 • Notice that the initial values carry the subscript 0, whilst the final values do not carry a subscript. A further simplification is achieved by choosing t0 =0 • Our equations now become: • vave = Δx/ Δt = (x – x0)/(t – t0) = (x – x0)/t • and aave = Δv/ Δt = (v – v0)/(t – t0) = (v – v0)/t • We restrict a to be constant so that aave = a and rearrange these equations to give: L4-s2,12

  3. Physics 114 – Lecture 4 • x – x0 = vave t • v = v0 + at • We need two further relations amongst these quantities • We note that these two equations both involve the time, t L4-s3,12

  4. Physics 114 – Lecture 4 • We now consider a plot of the velocity, v, as a function of the time, t. v v t Δt In time Δt distance travelled = v Δt = area shaded → total distance travelled = total area under v versus t curve. L4-s4,12

  5. Physics 114 – Lecture 4 • If a = constant the v versus t curve has the form: v at v0 t v0 t Distance travelled = total area under v versus t curve = area of shaded rectangle + area of shaded triangle = v0t + ½ t X at = v0t + ½ at2 L4-s5,12

  6. Physics 114 – Lecture 4 • We show that, if a = constant, then vave = (v + v0)/2 v E vave B A D at v0 v0 C t F t O Since a = const, the v versus t curve is a straight line and distance travelled = area under v versus t curve (blue line) = v0t + ½ at2 = area of CEFO = vaveх t = area of rectangle BDFO (red line). This is satisfiedif the area of triangle ABC = area of triangle ADE, which is, in turn, satisfied if vave = (v + v0)/2 L4-s6,12

  7. Physics 114 – Lecture 4 • With distance travelled = x – x0,we have • x – x0 = v0t + ½ at2 which we rearrange to give, • x = x0 + v0t + ½ at2 • We recall our two previous equations, • x – x0 = vave t • v = v0 + at • We note that each of these three equations involves the time, t L4-s7,12

  8. Physics 114 – Lecture 4 • and we also have, IF a =constant, • vave = (v + v0)/2 • We now need a relation involving a, v, v0, x and x0, but which does not include t • We shall eliminate t between the equations, • x – x0 = vave t • and v = v0 + at • and will also use the equation above, • vave = (v + v0)/2 L4-s8,12

  9. Physics 114 – Lecture 4 • Rearranging the equation, v = v0 + at • → at = v - v0 → t = (v - v0)/a • Using this value in x – x0 = vave t • → x – x0 = vave X (v - v0)/a • We now use vave = (v + v0)/2 to obtain, • x – x0 = [(v + v0)/2] X (v - v0)/a • → x – x0 = [(v + v0) X (v - v0)]/2a L4-s9,12

  10. Physics 114 – Lecture 4 • Reminder of algebraic relation, • a2 – b2 = (a + b) X (a – b) • Using this in x – x0 = [(v + v0) X (v - v0)]/2a • → x – x0 = (v2 - v02) /2a • Rearranging, 2a X (x – x0) = v2 - v02 • → v2 = v02 + 2a (x – x0) L4-s10,12

  11. Physics 114 – Lecture 4 • Collecting our four equations, we have • v = v0 + at no x, x0 • x = x0 + v0t + ½ at2 no v • v2 = v02 + 2a (x – x0) no t • vave = (v + v0)/2 and x – x0 = vave t no a • Notice that these relations hold only if a = constant L4-s11,12

  12. Physics 114 – Lecture 4 • Problem solving: • Determine which equation to use: the equation chosen must have only one unknown, all the other quantities being known. • Usually it is helpful if you can first determine t • The strategy is well summarized in Giancoli, p28 – the PROBLEM SOLVING box. • Lets look at some examples L4-s12,12

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