Asked by John
An object is moving in a straight line according to s(t)= -t^2+10t+30 where s(t) is how far the object goes in miles in t time in hours.
a. What is the constant acceleration of the object?
b. What is the farthest distance the object travelled from the starting point?
a. What is the constant acceleration of the object?
b. What is the farthest distance the object travelled from the starting point?
Answers
Answered by
merry christmas
idk r u talking in English
Answered by
oobleck
s(t)= -t^2+10t+30
v(t) = -2t+10
a(t) = -2
Now you can answer the questions.
v(t) = -2t+10
a(t) = -2
Now you can answer the questions.
Answered by
John
So the constant acceleration is a(t)= -2?
Answered by
oobleck
correct
Answered by
John
ok. so for my question b. would I plug in the acceleration into some equation to figure out farthest distance objecy traveled?
Answered by
oobleck
no. You want the maximum of s(t) which occurs when v(t) = 0
v = -2t+10
v=0 when t=5
s(5) = 55
Don't forget your Algebra I now that you are taking calculus. You just want to find the vertex of a parabola. You know that for y = ax^2+bx+c that occurs when x = -b/2a
Coincidentally, since dy'dx = 2ax+b, that is where y'=0
v = -2t+10
v=0 when t=5
s(5) = 55
Don't forget your Algebra I now that you are taking calculus. You just want to find the vertex of a parabola. You know that for y = ax^2+bx+c that occurs when x = -b/2a
Coincidentally, since dy'dx = 2ax+b, that is where y'=0
Answered by
John
Ok so the answer is 55 miles correct? Where are you getting the 5 from?
Answered by
oobleck
do you even read what I write?
v = -2t+10
v=0 when t=5
v = -2t+10
v=0 when t=5
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