Question

please solve question no 5 and 6

05.02.2 The acceleration- time data for a small rocket is given in tabular form below. 5. 20 22 18 24 16 14 12 10 Time (s) Acceleration 106.6 (m/s) 37.1 45.8 56.2 30.1 68.0 80.9 94.1 a) Use Newton's divided difference quadratic polynomial interpolation to find the acceleration at t = 15.5 seconds. Be sure to choose your hase points for good accuracy b) Use the quadratic interpolant of part (a) to find the change in the veloci of the rocket between t 14.1 and t = 15.8 seconds. of he A robot follows a path generated by a quadratic interpolant from x 2 to x = 4.The given by y x Find the length of the interpolant path from x 2 to x 4. You interpolant passes through three consecutive data points (2,4), (3,9) and (4,16) and is 6. can approximate a general integral by a+b (b-a) f(a) 6

Answer 1

Ques 5. A.

Newton divided difference quadratic polynomial interpolation

n	0	1	2	3	4	5	6	7
time (s)                            t	10	12	14	16	18	20	22	24
acceleration (m/s2)    f(t)	106.6	94.1	80.9	68	56.2	45.8	37.1	30.1
	-6.25	-6.6	-6.45	-5.9	-5.2	-4.35	-3.5
	-0.0875	0.0375	0.1375	0.175	0.2125	0.2125
	0.020833333	0.016667	0.00625	0.00625	-3E-16
	-0.000520833	-0.0013	9.25E-17	-0.00078
	-7.8125E-05	0.00013	-7.8E-05
	1.73611E-05	-1.7E-05
	-2.48016E-06

So, acceleration

a = 106.6 + (-6.25)*(t-10) + (-0.0875)*(t-10)(t-12) + 0.020833333*(t-10)(t-12)(t-14) + (-0.000520833)*(t-10)(t-12)(t-14)(t-16) + (-0.000078125)*(t-10)(t-12)(t-14)(t-16)(t-18) + 0.0000173611*(t-10)(t-12)(t-14)(t-16)(t-18)(t-20) + (-0.00000248016)*(t-10)(t-12)(t-14)(t-16)(t-18)(t-20)(t-22)

at t =15.5 sec,

a =106.6+(-6.25)*(15.5-10)+(-0.0875)*(15.5-10)*(15.5-12)+0.020833333*(15.5-10)*(15.5-12)*(15.5-14)+(-0.000520833)*(15.5-10)*(15.5-12)*(15.5-14)*(15.5-16)+(-0.000078125)*(15.5-10)*(15.5-12)*(15.5-14)*(15.5-16)*(15.5-18)+0.0000173611*(15.5-10)*(15.5-12)*(15.5-14)*(15.5-16)*(15.5-18)*(15.5-20)+(-0.00000248016)*(15.5-10)*(15.5-12)*(15.5-14)*(15.5-16)*(15.5-18)*(15.5-20)*(15.5-22) = 71.14145 m/s² Ans.

Ques 5. B.

acceleration = rate of change of velocity = dV/dt

so , dV = a dt

dV = -(0.00000031002*t⁸ – 0.000042162717*t⁷ + 0.00247251253*t⁶ - 0.08147572236*t⁵ + 1.64358557511*t⁴ - 20.70109380576*t³ + 161.6701015812*t² -843.400172336*t) + C

at t=15.8 sec , v = -2620.23374 + C

at t= 14.1 sec , v = -2764.17630 + C

Change in velocity of rocket between 14.1 sec to 15.8 sec = (-2620.23374 + C) – (-2764.17630 + C) = 143.94256 m/s