The velocity of a particle constrained to move along the x-axis as a function of time t is given by:

Question

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.

Answer 1

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

Answer 2

@Damon I believe you forgot a negative for the velocity. It should be -(15/t_0) sin(t/t_0).

Answer 3

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.