a) a = -3, b = 18, c = 48; s(t) = -3t²+18t+48b) 48mc) 8secs
Step-by-step explanation:
The question is incomplete. Here is the complete question.
'A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)=at²+bt+c, where t is the time in seconds after the rock was released. After 1 second the rock was 63 m above sea level, after 2 seconds 72 m, and after 7 seconds 27 m. a. Find a, b and c and hence an expression for s(t). b. Find the height of the cliff. c. Find the time taken for the rock to reach sea level.'
Given the equation of the distance modelled as s(t)=at²+bt+c
If after 1 second the rock was 63 m, then 63 = a+b+c
If after 2 seconds, the distance was 72 m then 72 = 4a+2b+c
Also if after 7 seconds, the distance is 27 m, then 27 = 49a+7b+c
i) Solving the 3 equations simultaneously to get a, b and c we have;
a+b+c = 63 ... 1
4a+2b+c = 72 ...2
49a+7b+c = 27 ...3
Subtracting 2 from 1 and 3 from 2 we will generate 2 new equations as shown;
eqn 2- eqn1: 3a + b = 9...4
eqn 3- eqn 2: 45a + 5b = -45
eqn 3- eqn 2: 9a+b = -9 ... 5
solving 4 and 5 simultaneously
3a + b = 9 ...4
9a+b = -9 ... 5
Taking the difference of 4 and 5 we have
6a - -9-9
6a = -18
a = -3
substituting a = -3 into equation 4 to get b we have;
3(-3)+b = 9
-9 + b = 9
b = 9+9
b = 18
substituting a = -3 and b = 18 into equation 1 to get c we have;
-3+18+c = 63
15+c = 63
c = 48
a = -3, b= 18 and c = 48
The distance function will be s(t) = -3t²+18t+48
ii) If the height of the cliff is modelled by the equation s(t)=at²+bt+c
The height of the cliff is at when t = 0
s(0) = -3(0)²+18(0)+48
s(0) = 48m
The height of the cliff is 48m
iii) At the sea level, the height of the rock will be 0m, substituting this into the modeled equation for the height to get the time we have;
s(t)=at²+bt+c
0 = -3t²+18t+48
3t²-18t-48 = 0
t² - 6t - 16 =0
t² - 8t+2t - 16 = 0
t(t-8)+2(t-8) = 0
(t+2)(t-8) = 0
t = -2 or 8
Taking the positive value of the time, t = 8secs
Time taken for the rock to reach sea level is 8secs
