The location of an object moving along the x-axis is given by the function x=10t+5t2−0.8t3 . At what positive time does the object stop moving?

Answer 1

To find the time at which the object stops moving, we need to find the value of t when the velocity of the object is equal to zero.

The velocity of an object is the derivative of its position function. So, let's first find the derivative of the given function:
x = 10t + 5t^2 - 0.8t^3

Differentiating both sides with respect to t, we get:
dx/dt = 10 + 10t - 2.4t^2

Now, set dx/dt equal to zero and solve for t:
10 + 10t - 2.4t^2 = 0

Rearranging the equation, we have:
2.4t^2 - 10t - 10 = 0

This equation is a quadratic equation, and we can solve it using the quadratic formula.
t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 2.4, b = -10, and c = -10.

Plugging in the values, we get:
t = (-(-10) ± √((-10)^2 - 4(2.4)(-10))) / (2(2.4))
t = (10 ± √(100 + 96)) / 4.8
t = (10 ± √196) / 4.8
t = (10 ± 14) / 4.8

Now, we have two possible solutions:
t = (10 + 14) / 4.8 ≈ 4.083
t = (10 - 14) / 4.8 ≈ -0.833

Since we're looking for a positive time, the object stops moving at t ≈ 4.083.