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?
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.
To find the positive time at which the object stops moving, we need to determine when the object's velocity is zero.
The velocity of an object can be obtained by taking the derivative of the position function with respect to time (t).
Given that x = 10t + 5t^2 - 0.8t^3, we can find the velocity function by differentiating with respect to t:
v = dx/dt = d/dt (10t + 5t^2 - 0.8t^3)
Taking the derivative of each term, we get:
v = 10 + 10t - 2.4t^2
Setting the velocity equal to zero and solving for t:
0 = 10 + 10t - 2.4t^2
Rearranging the equation, we have:
2.4t^2 - 10t - 10 = 0
This is a quadratic equation. We can solve it by factoring, completing the square, or by using the quadratic formula. In this case, let's solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from our equation:
t = (-(-10) ± √((-10)^2 - 4(2.4)(-10))) / (2(2.4))
Simplifying:
t = (10 ± √(100 + 96)) / 4.8
t = (10 ± √196) / 4.8
t = (10 ± 14) / 4.8
The two possible values for t are:
t1 = (10 + 14) / 4.8 ≈ 4.08
t2 = (10 - 14) / 4.8 ≈ -0.83
Since we are looking for the positive time when the object stops moving, we can conclude that the object stops moving at t = 4.08 seconds.
To find the time when the object stops moving, we need to determine when its velocity becomes zero.
The velocity of the object is given by the derivative of the equation for the location. Let's find the derivative of x with respect to t:
x = 10t + 5t^2 - 0.8t^3
Taking the derivative of both sides with respect to t:
dx/dt = d/dt(10t + 5t^2 - 0.8t^3)
= 10 + 10t - 2.4t^2
Set the derivative equal to zero and solve for t:
10 + 10t - 2.4t^2 = 0
Rearranging the equation:
2.4t^2 - 10t - 10 = 0
Now we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2.4, b = -10, and c = -10.
Plugging 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
Simplifying further:
t = (10 ± 14) / 4.8
So the two possible values for t are:
t1 = (10 + 14) / 4.8 = 24 / 4.8 = 5
t2 = (10 - 14) / 4.8 = -4 / 4.8 = -0.8333
Since we are looking for a positive time, the object stops moving at t = 5.