Question

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 5 sin πt + 2 cos πt, where t is measured in seconds. (Round your answers to two decimal places.)

(a) Find the average velocity during each time period.

(i) [1, 2]

? cm/s

(ii) [1, 1.1]

? cm/s

(iii) [1, 1.01]

?cm/s

(iv) [1, 1.001]

?cm/s

(b) Estimate the instantaneous velocity of the particle when t = 1

? cm/s

Please show all work

Answer 1

Given equation S - 5 sentit +2 Cos nit a) ) [19] nverage velocity on the interval (9,6] es (b)-SCO) - Vaverage : ba (12) Vaverage S(2)-5C1) paverage: 2.1 : (5 sin (291)+2 Cos (371))-(5 sin(n)+2 COS(TI)) : 500+2012-(500)+2(-)) : 0+2-10-2) - 2+2 Average velocity : 4 cmls : 4.00 cm/s (1, 1:1] nuerage velocity - S(1-1)-5(1) 1.1-1 :(55on (1171) + 3 COSCI.ITI))-(5sonen)+9 03 (1) 0.1 = 5(-0.309)+26-0.95106)-(500)+36-1)) 0.1 ; -3.44719-0+2 0.1 Average velocity = -14.47 cm/s

is 19,1.013 nuerage velocity: 5C7:00-5002 · 588n (1.011)+2008 (1-0111) . (sin(10) + 2008 1) 0.01 • 56-0.031411)+26-0.99951)-(5001426-1] 0.01 : -2.1561-0+2 0.01 nuerage velocity : -4561Cm/s --15.61cm/s. [1,1-0017 Average velocity : 501.001)-501) 1.001-1 Yu] · 5sen (1.00177)+2003(1.001TI) - [5 Sen (TI) + DOOS(TUJ] 0.001 = 5(-0.0031416 )+26-0.999995)-1510)+36-1)] 0.001 : -0.015698 -0+ 9 0.001 - 15-698 Average velocity .-15.70 cm/s b) Instantaneous velocity when t: ! S: 5 sentit +20 osat v(t): 45 = 5 coscrit). TI +2 (-sentit ) ? NCH) : 50coscrit)-20 senenti Vel) - 51icos (TTX1D-a91 Sin(Tix!

NI): 511 COS(11)-971 Sin(TV) ST(-1)-27(0) V(1): -511 :.instantaneous velocity : -Sticm/sec