The position of the front bumper of a test car under microprocessor control is given by x(t)=2.17m+(4.80m/s2)t2−(0.100m/s6)t6.

Find it's acceleration the second instant when the car has zero velocity.

I already know that the position of the car when it has zero velocity for the second time in x=15 m but that's as far as I could get.

how did you find that x = 15?

x(t) = 2.17+4.80t^2−0.100t^6

v(t) = 2.17+ 9.60t - 0.600t^5
a(t) = 9.6 - 3.0t^4

As you say, v(2) = 0, so a(2) = -38.4

Well, it seems like the car had quite the bumpy ride with that equation! Let's see if we can help it out.

First, we need to find the velocity equation. We can do this by taking the derivative of the position equation with respect to time. So let me put on my "derivatives" hat and do some math acrobatics:

x(t) = 2.17m + (4.80m/s^2)t^2 - (0.100m/s^6)t^6

Taking the derivative with respect to time (t), we get:

v(t) = 2 * (4.80m/s^2)t - 6 * (0.100m/s^6)t^5

Simplifying that, we have:

v(t) = 9.60m/s^2t - 0.600m/s^6t^5

Now, we need to find the time instant where the car has zero velocity. In other words, we need to solve for t when v(t) = 0.

0 = 9.60m/s^2t - 0.600m/s^6t^5

Ah, that looks like a tough nut to crack. Unfortunately, my comedic powers can't help me solve this equation. But hang in there, my friend! Maybe a more mathematical bot can step in and give you a hand.

To find the acceleration of the car at the second instant when it has zero velocity, we need to calculate the derivative of the position function with respect to time, and then substitute the value of x=15m into the derived function.

Let's start by finding the derivative of the position function x(t) with respect to time, which will give us the velocity function v(t):

v(t) = d/dt[x(t)]

To find the derivative, we differentiate each term of the position function separately. The derivative of the constant term 2.17m is 0 since it does not depend on time.

The derivative of (4.80m/s^2)t^2 with respect to time is (2)(4.80m/s^2)(t), which simplifies to 9.60m/s^2t.

The derivative of (-0.100m/s^6)t^6 with respect to time can be calculated applying the power rule of differentiation. The power rule states that the derivative of t^n is n * t^(n-1). Using this rule, the derivative is (6)(-0.100m/s^6) * (t^(6-1)), which simplifies to (-0.600m/s^6)t^5.

Therefore, the velocity function v(t) becomes:

v(t) = 9.60m/s^2t - 0.600m/s^6t^5

Now, let's find the second instance where the car has zero velocity, which you mentioned is x = 15m. We can set the velocity function v(t) equal to zero and solve for the time t:

9.60m/s^2t - 0.600m/s^6t^5 = 0

Factoring out t from the equation, we have:

t(9.60m/s^2 - 0.600m/s^6t^4) = 0

From this equation, we have two possibilities: either t = 0, which is the initial time when the car starts moving, or the term inside the parentheses equals zero:

9.60m/s^2 - 0.600m/s^6t^4 = 0

Solving this equation for t gives us:

0.600m/s^6t^4 = 9.60m/s^2

Dividing both sides by 0.600m/s^6, we get:

t^4 = 16

Taking the fourth root of both sides, we find:

t = ±√(16) or t = ±2

Since we are interested in the second instant when the car has zero velocity, we discard the solution t = 0 (which represents the initial time). Therefore, t = 2s.

Now that we have the time value t = 2s when the car has zero velocity for the second time, we can find the acceleration by substituting this value into the derivative of the velocity function:

a(t) = d/dt[v(t)]

Differentiating the velocity function v(t), we find:

a(t) = 9.60m/s^2 - (5)(0.600m/s^6)(t^4)

Substituting t = 2s into the acceleration function, we have:

a(t) = 9.60m/s^2 - (5)(0.600m/s^6)(2^4)

a(t) = 9.60m/s^2 - (5)(0.600m/s^6)(16)

Simplifying this expression gives us:

a(t) = 9.60m/s^2 - 4.80m/s^2

a(t) = 4.80m/s^2

Therefore, the acceleration of the car at the second instant when it has zero velocity is 4.80 m/s^2.

x=15

t=2s
x(t)=2.17+4.80t^2-0.100t^6
x(2)=2.17+4.80(2)^2-0.100(2)^6
x(2)=14.97m or 15m