An object at the origin at time t= 0 has velocity measured in meters per second,
v(t) = t/30 if 0 <= t <= 90
= 3 if 90 < t <= 108
= 9-(t/30) if 180 < t
When, if ever, does the object return to the origin? t=
s(t)
= t^2/15 if 0 <= t <= 90
= 90^2/15 + 3(t-90) if 90 < t <= 108
= 594 + 9(t-108) - (t-108)^2/15 if 180 < t
s(t) = 0 when t=291.54
Well, this object seems to have a bit of a wild ride with its velocity. Let's see if it ever finds its way back to the origin.
To determine when the object returns to the origin, we need to find when its position is also zero. The object's position can be found by integrating its velocity function.
For the first portion of the function, when 0 <= t <= 90, we can integrate v(t) = t/30. Integrating gives us s(t) = (1/60)t^2 + C, where C is a constant.
Now, to find the constant value C, we can use the fact that the object is at the origin at t = 0. Plugging this into our position equation, we get s(0) = 0, which means C = 0.
So our new position equation becomes s(t) = (1/60)t^2.
Now, let's look at the other portions of the velocity function. From 90 < t <= 108, the velocity is a constant 3. Integrating this gives us a position equation s(t) = 3t + C.
Again, to find the constant C, we can use the fact that at t = 90, the object's position is given by s(90) = (1/60)(90)^2. Plugging this into our equation and solving for C, we find C = -270.
So our new position equations are:
For 0 <= t <= 90: s(t) = (1/60)t^2
For 90 < t <= 108: s(t) = 3t - 270
Now, for the final portion of the velocity function, when t > 180, the velocity is given by v(t) = 9 - (t/30). Integrating this gives us s(t) = 9t - (1/60)t^2 + C.
To find the constant C, we can use the fact that s(108) = 3(108) - 270, since that's where the previous portion of the function left off at t = 108. Solving for C, we find C = 324 - 270 = 54.
So our final position equation, for t > 180, is s(t) = 9t - (1/60)t^2 + 54.
Now, we need to find when the object returns to the origin, which means when s(t) = 0.
For the equation s(t) = (1/60)t^2, the object never leaves the origin. So we don't need to consider that portion.
For the equation s(t) = 3t - 270, the object returns to the origin when 3t - 270 = 0. Solving for t, we find t = 90.
Finally, for the equation s(t) = 9t - (1/60)t^2 + 54, the object returns to the origin when 9t - (1/60)t^2 + 54 = 0. Unfortunately, this equation does not have any real solutions. So the object never returns to the origin in this portion.
Putting it all together, the object returns to the origin at t = 90.
To determine when the object returns to the origin, we need to find the time at which its position equals zero.
Let's first find the expression for the position function, s(t), by integrating the velocity function, v(t), with respect to time.
For 0 <= t <= 90:
v(t) = t/30
Integrating v(t), we get:
s(t) = ∫(t/30) dt = (1/2)(t^2/30) + C1
For 90 < t <= 108:
v(t) = 3
Integrating v(t), we get:
s(t) = ∫(3) dt = 3t + C2
For t > 180:
v(t) = 9 - (t/30)
Integrating v(t), we get:
s(t) = ∫(9 - (t/30)) dt = (9t - (1/2)(t^2))/30 + C3
Now, we can determine the constant terms C1, C2, and C3 using the initial condition that the object starts at the origin (s(0) = 0):
For 0 <= t <= 90:
s(0) = (1/2)(0^2/30) + C1 = C1 = 0
For 90 < t <= 108:
s(0) = 3(0) + C2 = C2 = 0
For t > 180:
s(0) = (9(0) - (1/2)(0^2))/30 + C3 = C3 = 0
Now, we can rewrite the position function including the determined constants:
For 0 <= t <= 90:
s(t) = (1/2)(t^2/30)
For 90 < t <= 108:
s(t) = 3t
For t > 180:
s(t) = (9t - (1/2)(t^2))/30
Now let's set the position function equal to 0 and solve for t to find when the object returns to the origin:
For 0 <= t <= 90:
(1/2)(t^2/30) = 0
t^2 = 0
t = 0
For 90 < t <= 108:
3t = 0
t = 0
For t > 180:
(9t - (1/2)(t^2))/30 = 0
9t - (1/2)(t^2) = 0
t(9 - (1/2)t) = 0
Solving the equation t(9 - (1/2)t) = 0, we find two solutions:
t = 0 and t = 18
Therefore, the object returns to the origin at t = 0 and t = 18.
To determine when the object returns to the origin, we need to find the values of time (t) at which the position (distance from the origin) of the object is zero.
The position at any time (t) can be found by integrating the velocity function over the given interval(s). Since the velocity function is divided into different intervals, we need to integrate separately for each interval.
For the first interval (0 <= t <= 90), the position function (x) can be found by integrating v(t) = t/30:
∫ (t/30) dt = (1/60) * t^2 + C1
Where C1 is an integration constant.
For the second interval (90 < t <= 108), the position function (x) can be found by integrating v(t) = 3:
∫ 3 dt = 3t + C2
Where C2 is an integration constant.
For the third interval (180 < t), the position function (x) can be found by integrating v(t) = 9 - (t/30):
∫ (9 - (t/30)) dt = 9t - (1/60) * t^2 + C3
Where C3 is an integration constant.
To find when the object returns to the origin, we need to solve the position functions for x = 0 in each interval and determine the corresponding values of time (t).
In the first interval:
(1/60) * t^2 + C1 = 0
Since t = 0 at the start of the interval, we can ignore the integration constant. Solving for t:
t^2 = 0
The only solution is t = 0.
In the second interval, the position function is linear: 3t + C2 = 0. Solving for t:
3t = -C2
t = -C2/3
In the third interval:
9t - (1/60) * t^2 + C3 = 0
We need to solve this quadratic equation for t. However, because this interval is open-ended (t > 180), we cannot determine an exact solution within this context. The solution would depend on the constant C3.
Therefore, the only time when the object returns to the origin is at t = 0 (at the start).