Question

Please explain reasoning and show all conversions and units for each equation, Thanks

The position of a particle moving along the x axis is given in centimeters by x = 9.75 + 1.50t3, where t is in seconds. Calculate (a) the average velocity during the time interval t = 2.00 s to t = 3.00 s; (b) the instantaneous velocity at t = 2.00 s; (c) the instantaneous velocity at t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s. (f) Graph x versus t and indicate your answers graphically.

Answer 1

In this question given id=9.75+ 1,50$? worth write in the form of atb u atbt3 since unit of a ts meter then unit of a ss meter . & unit of bt3 = m. b= m/ sec,

Sol Given x=9.75 +1.5043 a average relocity= total displacement - total time position of partical at t=2sec. Xx= 9.75*43+1,5 (2) 3 . = 9.75 + 12.0 = 21.75 & position of partical at st=3 sec X3 = 9.75 +6.5) (3) 50.25. then 50.25-21.75 average relocity = 3-2 : = 28.5 mis. b) 2 = 9.75+1.5043, = V = 3x1.50x+2. = 4.5+2 VH=2.cosec = 4.5*(2)? = 4.5 X4 = 18.0 m/s c) Vx=3sec = 4.5X(3) = 4.5*9. = 40.5 mis : ca) V4=2.5Sex =: 4.5*(25)? = 28.125 gmis.

(e) from a part of the question 22= 21.75. X3 = 50.25... mid position of X2 & Xz . = X27X3_21.75+50-25 2 : 2 = 72.00 = 36 m 2 of particle let at a sec. the position is 36m... 9:75.1.5+3 = 36 . 11573-26.25 .: +3= 17.5. S t = 2.6 sec. N2.6 = 4.5X1206)2 - 30.42 m/s No 9.75 +1.5%. de 9.75 A- 3

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