Question

An object moves such that its position on the x-axis is given by the function x(t)=a+bt+ct², where t is time. Given that x(1)=10, x(2)=11, and x(3)=2 find the coefficients a, b, and c.

(Linear Algebra Question)

Answer 1

Given that x(t)= a + b + ct2 and x (4) = 10, X62) - 11, (x(3)=2. Given x ct) = at bttct² put x c1)=10, le tot in -cij So, a+b+c= 90 - CA) put x (2)=11, c.e. t=2 in -ci) So, at2 b + 4C = 11 - (B) put x (3)=2 c.e. t=3 in – ci) So, at3b+90 = 2 -co) Now, (B)- A) gives b+30= – co) C)-CB) gives bt 50 = -9 – CE) Now, (o)-(E) gives c=25 put (205 in CD) b=16. pul (2-5, b=16 in (A) we get a = Thus, coefficients are -1 (a= -1, b = 16, C2-5 Ang