a = 2 t + 2
so
v = t^2 + 2 t + constant
-3 = 0 + 0 + constant
so
v = t^2 + 2 t -3
x(t) = (1/3)t^3 + t^2 - 3 t + constant
let x(0) = 0 so
x(t) = (1/3)t^3 + t^2 - 3 t
at t = 5
x = (1/3)(125) + 25 - 15 = indeed 155/3
= 51 2/3
HOWEVER v was - at start. It asked for the distance, not the displacement vector
when was v = 0?
v = t^2 + 2 t -3
0 = t^2+2t -3
(t+3)(t-1) = 0
t = 1
so how far back?
x(t) = (1/3)t^3 + t^2 - 3 t + constant
let x(0) = 0 so
x(3) = (1/3) + 1 - 3 = -5/3
so before it went up to x = 155/3
it went back 5/3 and forward 5/3
so total moved =155/3 + 10/3 = 165/3
=53
The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.
a(t) = 2t + 2, v(0) = −3, 0 ≤ t ≤ 5
(a) Find the velocity at time t.
v(t) = t^2+2t-3 m/s
(b) Find the distance traveled during the given time interval.
_____ m
I got B as 155/3. However my homework system won't take that answer and it doesn't want it as a decimal. Is that the answer for B or is there something else? It said A was right. Thank you.
1 answer
1 answer