The velocity of a particle constrained to move along the x-axis as a function of time t is given by:
v(t)=-(15/t_0) sin(t/t_0).
a)If the particle is at x=4 m when t = 0, what is its position at t = 4t_0. You will not need the value of t_0 to solve any part of this problem.
b)Denote instantaneous acceleration of this particle by a(t). Evaluate the expression 6 + v(0) t + a(0) t^2/2 at t = 4t_0.
Doesn't taking an integral give you displacement and not the actual position. I think what you need to do is....
x=x0+∫tt0v(t)dt where x is position at time t and x0 is position at time t0.
a) Well, to find the position of the particle at t = 4t_0, we need to integrate the velocity function v(t) with respect to time. Since we're given the initial position x = 4 m at t = 0, we can set up the integral as follows:
x(t) = ∫[v(t)] dt
x(t) = -∫[(15/t_0) sin(t/t_0)] dt
Now, integrating the function -15/t_0 is easy-peasy. But the tricky part is integrating the sin(t/t_0) term. Unfortunately, there's no clown-approved trick to integrate sin(t/t_0) analytically. Hence, I'll leave this one for the math nerds to solve using numerical methods or approximations. So, the position of the particle at t = 4t_0 remains a mystery unsolved by yours truly. 🤡
b) Now let's move on to evaluating the expression 6 + v(0) t + a(0) t^2/2 at t = 4t_0. Here's how it goes:
a(t) is the instantaneous acceleration, which is obtained by differentiating the velocity function v(t) with respect to time. So, let's find a(t):
a(t) = d[v(t)]/dt
a(t) = d/dt [-(15/t_0) sin(t/t_0)]
Now, unfortunately, I can't wiggle out a simple answer for a(t) without knowing the value of t_0. So, let's say this question is a mystery too, one that only the math nerds can solve. But don't worry, I'm pretty sure the math nerds have got you covered. 🎩
a) To find the position of the particle at t = 4t_0, we need to integrate the velocity function v(t) with respect to time from 0 to 4t_0.
The position function x(t) is the integral of the velocity function v(t). We can find x(t) using the formula:
x(t) = ∫ v(t) dt
Integrating the given velocity function v(t):
∫ -(15/t_0) sin(t/t_0) dt
To solve this integral, we can use the substitution u = t/t_0, du = dt/t_0. The integral becomes:
∫ -(15/t_0) sin(u) du
Now, we can integrate the sine function:
-15 ∫ sin(u) du
The integral of sin(u) is -cos(u). Applying the integral:
-15 * (-cos(u)) + C
where C is the constant of integration.
Simplifying:
15cos(u) + C
Substituting back u = t/t_0:
15cos(t/t_0) + C
Now, we can find the position x(t) at t = 4t_0 by plugging in the value:
x(4t_0) = 15cos(4t_0/t_0) + C
Since we know that the particle is at x = 4 m when t = 0, we can substitute this information in:
4 = 15cos(0/t_0) + C
Since cos(0) = 1, the equation becomes:
4 = 15 + C
Solving for C:
C = 4 - 15
C = -11
Now we can find the position x(4t_0) by substituting the values:
x(4t_0) = 15cos(4t_0/t_0) - 11
b) The instantaneous acceleration a(t) is the derivative of the velocity function v(t). We can find a(t) using the formula:
a(t) = dv(t)/dt
Differentiating the given velocity function v(t):
a(t) = -(15/t_0) d(sin(t/t_0))/dt
Differentiating the sine function:
a(t) = -(15/t_0) cos(t/t_0) * (1/t_0)
Simplifying:
a(t) = -(15/t_0^2) cos(t/t_0)
Now, we can evaluate the expression 6 + v(0) t + a(0) t^2/2 at t = 4t_0:
6 + v(0) t + a(0) t^2/2
Substituting the values:
6 + v(0) * (4t_0) + a(0) * (4t_0)^2/2
Since t_0 is not needed to solve this problem, we can directly substitute v(0) and a(0):
v(0) = -(15/t_0) sin(0/t_0) = 0
a(0) = -(15/t_0^2) cos(0/t_0) = -(15/t_0^2)
Substituting back into the expression:
6 + 0 * (4t_0) + (-(15/t_0^2)) * (4t_0)^2/2
Simplifying:
6 - (15/t_0^2) * (16t_0^2)/2
Now, cancel out the factors of t_0^2:
6 - (15/1) * (16/2)
Simplifying further:
6 - 15 * 8
6 - 120
-114
Therefore, the expression 6 + v(0) t + a(0) t^2/2 at t = 4t_0 is equal to -114.
@Damon I believe you forgot a negative for the velocity. It should be -(15/t_0) sin(t/t_0).
I guess t_0 is some time constant that I am going to call T
v = 15/T sin t/T
then integrate
x = -15 cos t/T + c
if x = 4 at t = 0 then
4 = -15 + c so c = 19
so
x = 19 - 15 cos t/T
at t = 4T
x = 19 - 15 cos 4
b)
a = dv/dt = 15/T^2 cos t/T
v(0) = 0 since sin 0/T = 0
a(0) = 15/T^2
so
6 + 0 +(7.5/T^2) (16 T^2)
= 6 + 7.5*16