Expert: Paul Klarreich - 6/21/2009

Question
What are the values of C_0 and C_1 in d(t) = C_1 + C_0 t - 16t^2, if d(1) = 4 and v(2) = -65?
A) C_0 = -1, C_1 = 21
B) C_0 = 1, C_1 = -21
C) C_0 = -1, C_1 = 19
D) C_0 = 0, C_1 = 1  

Answer
Questioner:   Blanca
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  What are the values of C0 and C1 in d(t) = C1 + C0 t - 16t^2, if d(1) = 4 and v(2) = -65?

A) C_0 = -1, C_1 = 21
B) C_0 = 1, C_1 = -21
C) C_0 = -1, C_1 = 19
D) C_0 = 0, C_1 = 1
...................................
Hi, again, Bianca,

If d(t) = C1 + C0 t - 16t^2, then  v(t) = C0 - 32t

d(1) = 4 means:

C1 + C0(1) - 16(1)^2 = 4, which means:

C1 + C0 - 16 = 4

C1 + C0 = 20  [A]

and v(2) = -65 means:

C0 - 32(2) = -65

C0 - 64 = -65

C0 = -1  [B]

Now you have two simultaneous equations A and B, and you already have C0.  Solve for C1:

C1 - 1 = 20  [A]

C1 = 21