An object is dropped from a height of 400ft.
(a) Find the velocity of the object as a function of time. (neglect air resistance) I think its t=400/v is that right??
(b) Use the result in (a) to find the position function.
(c) If air resistance is proportional to the square of the velocity, then dv/dt= -32+k(v)^2, where -32 fps is te acceleration dueto gravity and k is a constant. Show that the velocity v as a function of time is v(t)= -√ (32/k)tanh(√ (32k)t) by performing ∫dv/(32-k(v)^2)= - ∫dt and simplifying the result.
(d) Use the result of part (c) to find lim v(t) ( t to ∞) and give its interpretation.
(a) Find the velocity of the object as a function of time. (neglect air resistance) I think its t=400/v is that right??
(b) Use the result in (a) to find the position function.
(c) If air resistance is proportional to the square of the velocity, then dv/dt= -32+k(v)^2, where -32 fps is te acceleration dueto gravity and k is a constant. Show that the velocity v as a function of time is v(t)= -√ (32/k)tanh(√ (32k)t) by performing ∫dv/(32-k(v)^2)= - ∫dt and simplifying the result.
(d) Use the result of part (c) to find lim v(t) ( t to ∞) and give its interpretation.
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a. For this part, start with the acceleration (I'll choose down as negative acceleration) of -32 ft/s^2. If the object is dropped initial velocity is zero, and velocity is just the integral of acceleration, or
v(t) = -32t.
b. Position is the integral of velocity, and initial position is +400, so s(t) = -16t^2 + 400.
c. dv/dt = -32 + kv^2............multiply by dt, divide by (-32 + kv^2)
dv/(-32 + kv^2) = dt..............multiply both sides by -1
dv/(32 - kv^2) = - dt..............now you need a table of integrals; the left side will involve an inverse hyperbolic tangent, the right side is just - t. Rearrange, then take tanh of both sides to solve for v.
(I don't have a table handy. Sorry.)
d. The limit of v(t) is referred to as the terminal velocity of an object (max velocity when air resistance is considered).
v(t) = -32t.
b. Position is the integral of velocity, and initial position is +400, so s(t) = -16t^2 + 400.
c. dv/dt = -32 + kv^2............multiply by dt, divide by (-32 + kv^2)
dv/(-32 + kv^2) = dt..............multiply both sides by -1
dv/(32 - kv^2) = - dt..............now you need a table of integrals; the left side will involve an inverse hyperbolic tangent, the right side is just - t. Rearrange, then take tanh of both sides to solve for v.
(I don't have a table handy. Sorry.)
d. The limit of v(t) is referred to as the terminal velocity of an object (max velocity when air resistance is considered).
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a) ... you're looking for v (t)
acceleration due to gravity = - 32 ft / s^2
velocity is the integral of acceleration
v (t) = - 32 t <=== velocity as a function of time
b) ... position (s) is the integral of acceleration
. . . . given initial position = 400
s (t) = - 16 t^2 + 400 <=== position as a function of time
acceleration due to gravity = - 32 ft / s^2
velocity is the integral of acceleration
v (t) = - 32 t <=== velocity as a function of time
b) ... position (s) is the integral of acceleration
. . . . given initial position = 400
s (t) = - 16 t^2 + 400 <=== position as a function of time