The height, y, of an object thrown into the air is known to be given by a quadratic function of t (time) of the form y = at2 + bt + c. If the object is at height y = 23/4 at time t = 1/2, at y = 7 at time t = 1, and at y = 2 at t = 2, determine the coefficients a, b, and c.
Answers
Answer:
a = -5
b = 10
c = 2
Step-by-step explanation:
y = at² + bt + c
object is at height y = 23/4 at time t = 1/2
=> 23/4 = a(1/2)² +b(1/2) + c
=> 23 = a + 2b + 4c Eq1
object is at height y = 7 at time t = 1
=> 7 = a + b + c Eq 2
object is at height y = 2 at time t = 2
=> 2 = 4a + 2b + c Eq 3
Eq1 - Eq 2
=> 16 = b + 3c
4 * eq2 - Eq3
=> 26 = 2b + 3c
=> 26 = b + b + 3c
=> 26 = b + 16
=> b = 10
16 = 10 + 3c
=> c = 2
a + b + c = 7
=> a + 10 + 2 = 7
=> a = -5
y = -5t² + 10t + 2
a = -5
b = 10
c = 2
Answer:
The values of a,b and c are -5, 10 and 2.
Step-by-step explanation:
The first three equations were :
1) 5.75=a*0.25 +b*0.5 +c
2) 7=a +b +c
3) 2=4*a +2*b +c
We have obtained these equations by simply substituting the data into the given expression. We first solved eq 3 and 2 to eliminate c and obtained eq 4 which is : 5=-3*a-b. We solved equation 1 and 2 to eliminate c and obtained eq 5 ,which is : -1.25=-0.75 *a-0.5*b. We solved equation 4 and 5 simultaneously and obtained the value of a which is -5. Then we substituted it in eq 4 and obtained b=10. Finally we substituted both these values in eq 2 and obtained the value of c which is 2.