clothinstylewas alsal ConrecutedefectoQingeststratedy thee citpulaiuropwaltyighEurrivinostanderitanvegove lS,fcktDIh1(PSET 2: Particle Motion-11 Questions to Think About Name:Due on Friday 12/02t0xa (t)va (t) -4723-15118531110113250156-6VParticle a and particle b are moving along a horizontal line for 0 st≤15. The velocity of particle a isgiven by v. (t) which is a continuous and twice differentiable function that only changes directiontwice. It is known that particle a has an average acceleration of zero over the interval 3 6. For what values of tis particle b speeding up? Justify.7. Find all values of m where the average acceleration of particle bon 3≤t≤m is zero.8. For what values of tis v(t)=0?9. What is the minimum number of times that x (t) must equal 7? Justify.10. Which particle has a greater average speed on 8≤t≤11? Justify.11. For what values of tare both particles moving to the right?

Name: clothinstylewas alsal ConrecutedefectoQingeststratedy thee citpulaiur
Uploaded: 2024-03-05T11:56:34.000Z
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Description: clothinstylewas alsal ConrecutedefectoQingeststratedy thee citpulaiuropwaltyighEurrivinostanderitanvegove lS,fcktDIh1(PSET 2: Particle Motion-11 Questions to Think About Name:Due on Friday 12/02t0xa (t)va (t) -4723-15118531110113250156-6VParticle a

Given:position xa, and velocity va of particle a at correspondingtimes is given.Also graph of velocity of particle b vs time is given.Solution:(Sign convention: Direction to the right is taken as positiveand direction to left is taken as negative.)1)We have a straight line at from t=0 to t=3swith y intercept, c = 0 thereforevb = ab×there ab is slope of the lineab = 2−03−0 = 23 m/s2this slope is the acceleration of the particle bfrom t=0 to 3shencevb = 23t for t=0 to t=3shenceat t= 2s vb = 23×2vb = 1.33 m/s2)as seen from question 1 ab = 23 m/s2 for t=0 to 3 shence for t= 2 s as wellab = 23 = 0.666 m/s23) Average velocity for time t1 to t2 is given byvavg = x(t2)−x(t1)t2−t1similarlyfor particle a, from time t= 2s to 8sva_avg = xa(8)−xa(2)8−2 = 5−36 = 26va_avg = 0.333 m/s4) For b to move to left its velocity must be negative from the graph, we see that vb is negative for 7.5 s<t<13 shence particle b is moving towards left in the time interval, 7.5 s <t < 13 s5) For the acceleration of b to be zerothe graph must have slope = 0 and since we want particle moving to right vb must be positiveWe see that the only time at which slope of the graph is zero and the is vb is positiveis at t= 5 sat top of semicircular region slope of vbvs t graph is momentarily zero hence from the graph vb at t=5 s is vb = 4 m/s6) By speeding up means that the slope of vbvs tgraph must be positive, or in other words acceleration must be postive.We see that slope is positive and hence particle b is speeding up for time intervals,a) from t=0 s to 5 s b) from t = 10 s to t=15 s7) Average acceletaion between two time pointst1 and t2 is given asaavg = v(t2)−v(t1)t2−t1for avg acceleration to be zero,aavg = 0= v(t2)−v(t1)t2−t1∴ v(t2)−v(t1) = 0hencev(t2) = v(t1)now given t1 = 3 sand t2 = m swe have vb(3 s) = 2 m/s (from the graph)also we see that at t=7 s and at t= 14 svb(t) = 2 m/shencem can have valuesm = 7 sm = 14 s8) We see clearly from the table for particle a that va = 0 m/s at t= 13 sBut also there is an instance between t=2 s to t = 5 sas at t=2 s , va is negative and at t=5 s the va is positive hencether must be a va = 0 m/s between t=2 s and t=5 sas va(t) is double differentiable, aa(t) will be time varying, hence finding exact value of t is not possible but va = 0 at some instance between t= 2 s and t=5 s9)we see that xa(t) = 7 at t= 0 sxa(t) transistins from 5 to 10 from t= 8 s to t = 11 shence in between 8 and 11 s there must be another instance at which xa(t) = 7 Also xa(t) transitions from 25 to 6 fro t= 13 s to 15 stherefore there must be another instance when xa(t) = 7 between t = 13 s to 15shence there are total 3 instances when xa(t) = 7 10)Average velocity of a for 8≤t≤11 is va_avg = xa11−xa811−8 = 10−53 = 53va_avg = 1.667 m/sAverage velocity of b for 8≤t≤11 is is equal to −2 m/s as the vbt is constant in the given time period.Now as we are asked speed, we are to takeonly magnitude in consideration,hencewe see that −2 > 1.667henceaverage speed of particle b is greater in the interval, 8≤t≤1111)From t= 5 to t=13, particle a is having positive velocity hence is moving to right In this interval we see that particle b has positivevelocity in interval t= 5 to t=7.5Hence in the interval 5≤t≤7.5 both particles are moving to the right