Question

And BTW I already tried entering it as 426 ft (3 sig figs).

Problem 12.14 < 3 of 9 A Revie The position of a particle along a straight-line path is defined by s = (t - 6t? – 15t + 7) ft, where t is in seconds Part A Determine the total distance traveled when t = 10.9 s. Express your answer to three significant figures and include the appropriate units. I HÅR O 2 ? ST = 425.669 Submit Previous Answers Request Answer * Incorrect; Try Again; 4 attempts remaining

Answer 1

AJE First thing we will do is to write the expression for velocily, in order to determine how many time, the particle changes direction, de haumany it speed as equal to zero the inestantanidny velocity actually average velocity Measured at infinitesimal intowal of time, so the expression for instaniay velocity is de morder to determine time when particle is still, We replace ubyo a so o=3t-127-15 the roots of the quadratic equation above are tyn = 1273 B) o tz al Given that the velocity is negative at t=0, ce(-157 We can say that the particle move to the back, then stops at t=5 and than moves to front. atenta we know s=²_6t15t+7 we will replace t by 0,5 and 6-9 to determine the position of the particle at those time s(t=0)= 03-6(6)-1510)+7 > School - 7 ft slt=5)=536157-15(5) +7 ass=-9344 s(t=1009) = 106(1897-15 10:9)+7 =425669.14 Now that we have all the details... CSdraw the path of the particle aus, we can correal Scanned with

we can see from the sketch that the total distance travelled is Ś= About AS-2 $=(5,-Solt(s.-s, I ST-17-(-93) 1 + 1 425667-693)) = 618.667 44 5,2-934 11 ---- Paol 12 t2=5s < ---- Po -- - -V3 O P - - - -- - S3=625. 669 S = 6180669 Lt anned with ofal distance trowelled CamScanner