To find the time at which the object stops moving, we need to find the time when its velocity reaches zero. The velocity of the object is given by the derivative of the function x(t) with respect to t.
Taking the derivative of x(t) = 10t + 5t^2 - 0.8t^3 gives:
v(t) = 10 + 10t - 2.4t^2
Setting v(t) = 0 and solving for t:
10 + 10t - 2.4t^2 = 0
Rearranging the equation:
2.4t^2 - 10t - 10 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 2.4, b = -10, and c = -10. Substituting these values into the quadratic formula:
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
This gives us two possible values for t:
t1 = (10 + 14) / 4.8 = 24 / 4.8 = 5
t2 = (10 - 14) / 4.8 = -4 / 4.8 = -0.833
Since we are looking for a positive time, the object stops moving at t = 5.
Therefore, the correct response is:
t = 5.0
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? (1 point) Responses t = 0.8 t = 0.8 t = 1.6 t = 1.6 t = 3.2 t = 3.2 t = 5.0 t = 5.0 t = 10.0
1 answer