Free fall physics expression

[pringless]

An object is releaseed from rest at time t = 0 and falls through the air, which exerts a resistive force such that the acceleration a of the object is given by a = g - bv, where v is the object's speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time?

a) v = g(1-e^bt) / b
b) v = (g+a)t / b
c) v = (ge^bt) / b
c) v = gt - bt^2
d) v = v_0 + gt, v_0 does not = 0

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The velocity of a particle moving along the x-axis is given by:

v = at - bt^3 for t > 0

a = 31m/s^2, b = 2.5 m/s^4, and t is in seconds.

What is the acceleration a, of the particle when it achieves its maxiumum displacement in the positive x direction? Answer in units of m/s^2.

 

[HallsofIvy]

For the first problem,the main thing you need to think about is "will the speed get greater or less? Which of the possible answer gives that?


For the second problem, as long as the velocity is positive, the particle is still moving forward. When the velocity is negative, it's already going to the left. The largest value of x 0ccurs when the veocity is 0.

 

[M98Ranger]

The acceleration of the particle at its maximum displacement in the positive x direction can be found by taking the derivative of the velocity function with respect to time:

a = dv/dt = a - 3bt^2

At the maximum displacement, the velocity is 0, so we can set v = 0 and solve for t:

0 = at - bt^3
t = a/b

Substituting this value of t into the acceleration function, we get:

a = a - 3b(a/b)^2
a = a - 3a^2/b

Simplifying, we get:

a = a(1 - 3a/b)

Thus, the acceleration at the maximum displacement in the positive x direction is given by a(1 - 3a/b) m/s^2.