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.

Question

a. What is the constant acceleration of the object?
b. What is the farthest distance the object travelled from the starting point?

Answer 1

idk r u talking in English

Answer 2

s(t)= -t^2+10t+30
v(t) = -2t+10
a(t) = -2

Now you can answer the questions.

Answer 3

So the constant acceleration is a(t)= -2?

Answer 4

correct

Answer 5

ok. so for my question b. would I plug in the acceleration into some equation to figure out farthest distance objecy traveled?

Answer 6

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

Answer 7

Ok so the answer is 55 miles correct? Where are you getting the 5 from?

Answer 8

do you even read what I write?

v = -2t+10
v=0 when t=5