The velocity of a particle constrained to move along the x-axis as a function of time t is given by:
v(t)=-(13/t_0) sin(t/t_0).
a)If the particle is at x=4 m when t = 0, what is its position at t = 5t_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 4 + v(0) t + a(0) t^2/2 at t = 5t_0. You will find that this does not agree with the answer to part (a), which is expected because the acceleration in this problem is changing with time.
c)Assume that the particle is at x = 0 when t = 0, and find its position at t = 0.15t_0.
d)Evaluate v(0) t + a(0) t^2/2 at t =0.15t_0 to 3 significant figures. Your answers to parts (c) and (d) should be in good agreement. Try to work out why that is so before submitting the answer to part (d)
a) position(t)=position(t=0)+v(t)*t
I will be happy to critique your thinking.
I would know how to solve this problem if i were to understand the whole t/t_0 thing. I don't know how to integrate something like that. Could you please explain how to get it started
There is no integration here, it is straight calculation with your calculator.
so i get v(t)=-(13*5)/t
To solve these problems, we need to integrate the given velocity function, v(t), to obtain the position function, x(t). We will start by integrating v(t) with respect to t.
a) To find the position at t = 5t_0, we integrate v(t) over the interval [0, 5t_0]:
x(t) = ∫[0,5t_0] v(t) dt
Since v(t) = -(13/t_0) sin(t/t_0), the integral becomes:
x(t) = ∫[0,5t_0] -(13/t_0) sin(t/t_0) dt
To evaluate this integral, we can use the substitution u = t/t_0. This gives us:
x(t) = -13 ∫[0,5] sin(u) du
Now, we can integrate sin(u) with respect to u:
x(t) = -13 [-cos(u)] [0,5]
x(t) = 13 (cos(0) - cos(5))
Since cos(0) = 1 and cos(5) ≈ 0.283662, we get:
x(t) ≈ 13 (1 - 0.283662) = 10.683306 m
Therefore, the position of the particle at t = 5t_0 is approximately 10.683306 m.
b) To evaluate the expression 4 + v(0) t + a(0) t^2/2 at t = 5t_0, we need to find the instantaneous acceleration, a(t), and then substitute the values into the expression. Let's first find a(t) by taking the derivative of v(t) with respect to t:
a(t) = dv(t)/dt
Using the chain rule, we differentiate v(t) = -(13/t_0) sin(t/t_0):
a(t) = (13/t_0^2) cos(t/t_0)
Now, we can substitute t = 0 into v(0) and a(0) to get their values:
v(0) = -(13/t_0) sin(0/t_0) = 0
a(0) = (13/t_0^2) cos(0/t_0) = 13/t_0^2
Now, we substitute these values into the expression 4 + v(0) t + a(0) t^2/2 and evaluate it at t = 5t_0:
4 + v(0) t + a(0) t^2/2 = 4 + 0(5t_0) + (13/t_0^2)(25t_0^2)/2
= 4 + 0 + 13/2
= 4 + 6.5
= 10.5
The expression evaluates to 10.5, which does not agree with the result from part (a).
c) To find the position when the particle is at x = 0 and t = 0, we integrate v(t) over the interval [0, 0.15t_0]:
x(t) = ∫[0,0.15t_0] v(t) dt
Using the same process as in part (a), we find:
x(t) = 13(cos(0) - cos(0.15))
Since cos(0) = 1 and cos(0.15) ≈ 0.988771, we have:
x(t) ≈ 13(1 - 0.988771) = 0.159297 m
Therefore, the position of the particle at t = 0.15t_0 is approximately 0.159297 m
d) To evaluate v(0) t + a(0) t^2/2 at t = 0.15t_0, we substitute t = 0.15t_0 into v(0) and a(0):
v(0) = -(13/t_0) sin(0/t_0) = 0
a(0) = (13/t_0^2) cos(0/t_0) = 13/t_0^2
Now, we substitute these values into v(0) t + a(0) t^2/2 and evaluate it at t = 0.15t_0:
0 + (13/t_0^2)(0.15t_0)^2/2 = (13/t_0^2)(0.0225t_0^2)/2
= (13/2t_0^2)(0.0225t_0^2)
= 13 * 0.0225/2
≈ 0.146625
The expression evaluates to approximately 0.146625, which is in good agreement with the result from part (c).
The reason the answers to parts (c) and (d) are in good agreement is that the initial position and initial velocity are both zero in part (c). Therefore, the term v(0) t in part (d) cancels out, leaving only the term a(0) t^2/2, which matches the position calculation in part (c).