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.
To find the velocity function v(t), we can integrate the acceleration function a(t) with respect to time.
Given that a(t) = 2t + 2, we integrate it to obtain v(t).
∫(2t + 2) dt = t^2 + 2t + C
Since v(0) = −3, we can substitute t = 0 into the equation to find the constant C.
v(0) = 0^2 + 2(0) + C = -3
C = -3
Thus, the velocity function v(t) is:
v(t) = t^2 + 2t - 3 m/s
Now, to calculate the distance traveled during the given time interval (0 ≤ t ≤ 5), we need to integrate the absolute value of the velocity function over this interval.
∫|t^2 + 2t - 3| dt from 0 to 5
To solve this, we can break the integral into two parts:
∫(t^2 + 2t - 3) dt from 0 to 5 (when t^2 + 2t - 3 > 0)
and
∫-(t^2 + 2t - 3) dt from 0 to 5 (when t^2 + 2t - 3 < 0)
Simplifying the two integrals:
∫(t^2 + 2t - 3) dt = (1/3)t^3 + t^2 - 3t
∫-(t^2 + 2t - 3) dt = -(1/3)t^3 - t^2 + 3t
Evaluating the integrals:
[ (1/3)(5^3) + 5^2 - 3(5) ] - [(1/3)(0^3) + 0^2 - 3(0)] = 155/3
Therefore, the distance traveled during the given time interval is 155/3 meters.
If your homework system does not accept the answer as a fraction, you can convert it to a decimal. The decimal equivalent of 155/3 is approximately 51.67.
To find the distance traveled during the given time interval, we can use the formula:
d(t) = ∫[v(a), v(b)] v(t) dt
Given that 0 ≤ t ≤ 5, we need to integrate the velocity function from t = 0 to t = 5.
(a) The velocity function is given as v(t) = t^2 + 2t - 3 m/s.
(b) Now, we can calculate the distance using the formula:
d(t) = ∫[0, 5] (t^2 + 2t - 3) dt.
Evaluating the integral:
d(t) = (∫t^2 dt) + (∫2t dt) - (∫3 dt)
= (t^3/3 + t^2 - 3t) [0, 5]
= (5^3/3 + 5^2 - 3(5)) - (0^3/3 + 0^2 - 3(0))
= (125/3 + 25 - 15) - (0/3 + 0 - 0)
= 125/3 + 25 - 15
= 50/3 + 25
= (50 + 75)/3
= 125/3
The distance traveled during the given time interval is 125/3 meters.
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