Question

Given the accerleration, initial position of a body moving along a coordinate line at time t, find the bodys position at time t
a=120, v(0)=2,s(0)=-4

A s=60t^2 +2t-4
B s=120t^2 +2t-4
Cs= -60t^2-2t-4
D s=60t^ =2t

Can someone help

Answer 1

is this the whole question?

Answer 2

yes

Answer 3

its Cal 1

Answer 4

i believe you can use the formula \[s_2 - s_1 = v_1 t + \frac 12 at^2\]
where s2 is final position
s1 is initial position
v1 is inital speed
t = time
a = acceleration

Answer 5

plug in your values and tell me what you ahve

Answer 6

have*

Answer 7

i have 60t^2 + 2t is that right. But what do i do with the s(0)=-4?

Answer 8

hmm..if you tried plugging in the values into the formula i stated you would have gotten

s - (-4) = 2t + 1/2 (120) t^2

do you see it now?

Answer 9

so would it be s=60t^2 +2t -4?

Answer 10

yes

Answer 11

Thanks so much i appreciate your help!

Answer 12

you're welcome :DD

Answer 13

hey do you think you could help me with another ?

Answer 14

why not :D

Answer 15

Find the value or values of c that satisfy the equation f(b)-f(a)/b-a= f'(c) in the conclusion of the Mean Value Theroem for the function and interval. Round to the nearest thousandth

F(x)=tan^-1(x), [-sqrt(3), sqrt(3)]

Answer 16

awww not good with this sorry :(

Answer 17

i hated those theorems lol

Answer 18

its cool thank anyway i hate them too! lol