Motion: Velocity and Net Change

1 Math 131,applicationsmotion: Velocity and net Change 1 motion : Velocity and Net ChangeIn Calculus I you interpreted the first and second derivatives as Velocity and accel-eration in the context of motion . So let s apply the initial value problem results tomotion problems. Recall that s(t) =position at timet. s (t) =v(t) = Velocity at timet. s (t) =v (t) =a(t) =acceleration at a(t)dt=v(t) +c1= Velocity . v(t)dt=s(t) +c2=position at will need to use additional information to evaluate the that the acceleration of an object is given bya(t) =2 costfort 0 with v(0) =1, this is also denotedv0 s(0) =3, this is also (t). findv(t)which is the antiderivative ofa(t).

2 V(t) = a(t)dt= 2 cost dt=2t sint+ use the initial value forv(t)to solve forc1:v(0) =0 0+c1=1 c1= ,v(t) =2t sint+1. Now solve fors(t)by taking the antiderivative ofv(t).s(t) = v(t)dt= 2t sint+1dt=t2+cost+t+c2 Now use the initial value ofsto solve forc2:s(0) =0+cos 0+c2=3 1+c2=3 c2= (t) =t2+cost+2t+ acceleration is given bya(t) =10+3t 3t2, find the exact positionfunction ifs(0) =1 ands(2) = (t) = a(t)dt= 10+3t 3t2dt=10t+32t2 t3+ (t) = 10t+32t2 t3+c dt=5t2+12t3 14t4+ct+ (0) =0+0 0+0+d=1 sod=1. Thens(2) =20+4 4+2c+1=11 so2c= 10 c= 5. Thus,s(t) =5t2+12t3 14t4 5t+ acceleration is given bya(t) =sint+cost, find the position functionifs(0) =1 ands(2 ) = (t) = a(t)dt= sint+cost dt= cost+sint+ (t) = cost+sint+c dt= sint cost+ct+ (0) =0 1+0+0+d=1 sod=2.

3 Thens(2 ) =0 1+2c +2= 1 so2 c= 2 c= 1 +2. Thus,s(t) =cost+sint 1 131,applicationsmotion: Velocity and net Change 2 Displacement vs Distance an object between timest=aand a later timet=biss(b) s(a) = bav(t) other words,displacementis thenetareabetween the Velocity curve andthe horizontal axis, while thedistancetravelledis thetotal areabetween thevelocity curve and the horizontal travelledby an object between timest=aand a later timet=bisDistance travelled= ba|v(t)| an object moves with Velocity 2t2 12t+16 km/hr.(1) Determine the displacement of the object on the time interval[1, 3]and[0, 4]and interpret your answer.(2) Determine the distance travelled on[0, 4]SOLUTION.

4 (1) The displacement is easy to calculate: For the interval[0, 4], On[0, 4],s(4) s(0) = 402t2 12t+16dt=2t33 6t2+16t 40=1283 96+64 0= [1, 3],s(3) s(1) = 312t2 12t+16dt=2t33 6t2+16t 41= (18 54+48) (23 6+16)=43.(2) The distance travelled is harder to determine since we need to integrate|v(t)|.We must first determine wherev(t)is positive and 12t+16+ + + ++002t2 12t+16=2(t2 6t+8) =2(t 2)(t 4) =0 t=2, number line to the right shows that 2t2 12t+16 0 only[2, 4]. We cannow find the distance travelled (total area) by splitting the interval into twopieces[0, 2]and[2, 4], changing the sign ofv(t)on the second piece to obtainthe absolute value ofv(t).

5 Dist Trav=Total Area= 40|2t2 12t+16|dt= 202t2 12t+16dt+ 42 (2t2 12t+16)dt= [2t33 6t2+16t 20] [2t33 6t2+16t 42]=403+83= : The distance travelled on[0, 4]is the area under the absolutevalue of the Velocity an object moves with velocityt3 5t2+4tm/s.(1) Determine the displacement of the object on the time interval[0, 6]and interpretyour answer.(2) Determine the distance travelled on[0, 6]SOLUTION.(1) For displacement on[0, 6], 60t3 5t2+4t dt=t44 5t33+2t2 60= (324 360+72) 0=36.(2) For the distance travelled we must first determine wherev(t)is positive 146t3 5t2+4t+ ++0 00t3 5t2+4t=t(t2 5t+4) =t(t 1)(t 4) =0 t=0, 1, 131,applicationsmotion: Velocity and net Change 3 The number line to the right shows thatt3 5t2+4t 0 only[1, 4].

6 We can nowfind the distance travelled (total area) by splitting the interval into three pieces[0, 1],[1, 4]and[4, 6], changing the sign ofv(t)on the second piece to obtain theabsolute value ofv(t).Dist Trav= 60|2t2 12t+16|dt= 10t3 5t2+4t dt 41t3 5t2+4t dt+ 64t3 5t2+4t dt=[t44 5t33+2t2 10] [t44 5t33+2t2 41]+[t44 5t33+2t2 64]=712+454+1403= 2 3 4 5 6 Acceleration: GravityIn many motion problems the acceleration is constant. This happens when anobject is thrown or dropped and the only acceleration is due to gravity. In such asituation we have a(t) =a, constant acceleration with initial velocityv(0) =v0 and initial positions(0) = (t) = a(t)dt= a dt=at+ (0) =a 0+c=v0 c= (t) =at+ ,s(t) = v(t)dt= at+v0dt=12at2+v0t+ timet=0,s(0) =12a(0)2+v0(0) +c=s0 c= (t) =12at2+v0t+ a ball is thrown with initial Velocity 96 ft/s from a roof top432feet high.

7 The acceleration due to gravity is constanta(t) = 32 ft/s2. Findv(t)ands(t). Then find the maximum height of the ball and the time when the ball hits thatv0=96 ands0=432 and that the acceleration isconstant, we may use the general formulas we just (t) =at+v0= 32t+96ands(t) =12at2+v0t+s0= 16t2+96t+ 131,applicationsmotion: Velocity and net Change 4 The max height occurs when the Velocity is 0 (when the ball stops rising):v(t) = 32t+96=0 t=3 s(3) = 144+288+432=576 ball hits the ground whens(t) = (t) = 16t2+96t+432= 16(t2 6t 27) = 16(t 9)(t+3) = only (sincet= 3 does not make sense). person drops a stone from a bridge. What is the height (in feet) of thebridge if the person hears the splash5seconds after dropping it?

8 S what we (dropped) ands(5) =0 (hits water). Andwe know acceleration is constant,a= 32 ft/s2. We want to find the height of thebridge, which is justs0. Use our constant acceleration motion formulas to solve (t) =at+v0= 32tands(t) =12at2+v0t+s0= 16t2+ we use the position we know:s(5) = (5) = 16(5)2+s0 s0=400 that we did not need to use the Velocity TRY (Extra Credit).In the previous problem we did not take into account thatsound does not travel instantaneously in your calculation above. Assume that sound trav-Check on your answer: Should thebridge be higher or lower than in thepreceding example? Why?els at1120ft/s. What is the height (in feet) of the bridge if the person hears the splash5seconds after dropping it?

9 S a variation. This time we will use metric units. Suppose a ballis thrown with unknown initial velocityv0m/s from a roof top49meters high andthe position of the ball at timet=3 iss(3) =0. The acceleration due to gravity isconstanta(t) = m/s2. Findv(t)ands(t). timev0is unknown buts0=49 ands(3) =0. Again the accelerationis constant so we may use the general formulas for this (t) =at+v0= +v0ands(t) =12at2+v0t+s0= +v0t+ we know thats(3) = (3)2+v0 3+49=0which means3v0= (9) (10) = v0= (t) = 4930ands(t) = 4930t+ Green is attempting to run the100m dash in the Geneva Invita-tional Track Meet He wants to run in a way that hisaccelerationisconstant,a, over the entire race.

10 Determine his Velocity function. (awill still appear asan unknown constant.) Determine his position function. There should be no unknownconstants in your equation at this point. What is his Velocity at the end of the race?Do you think this is realistic?math 131,applicationsmotion: Velocity and net Change have: constant acceleration=am/s2;v0=0 m/s;s0=0 m. Sov(t) =at+v0=atands(t) =12at2+v0t+s0= ( ) =12a( )2=100, soa=200( )2= m/s2. Sos(t) = Mo svelocity at the end of the race isv( ) =a ( ) = m/s.. stone dropped off a cliff hits the ground with speed of120 was the height of the cliff? thatv0=0 (dropped!) ands0is unknown but is equal to the cliffheight, and that the acceleration is constanta= 32 ft/.