1. The current flowing through an electric circuit is the derivative of the charge as a function of time. If the charge q is given by the equation q(t)=3t^2+2t-5, what is the current at t=0?
a. -5
b. 0
c. 2
d. 8
2. The equation for the position of an object at time t is represented by the equation f(t)=4t^2-2t. Which equation represents the instantaneous velocity at any given time, t?
a. v(t)=8t
b. v(t)=4t
c. v(t)=4t-2
d. v(t)=8t-2
3. The position of a train as it travels between two stations is described by the equation f(t)=90t^2+2t, where f(t) is the distance in miles and t is time in minutes. What is its average velocity between minutes 3 and 7?
a. 816 mi./min.
b. 902 mi./min.
c. 1,310 mi./min.
d. 4,424 mi./min.
4. The velocity of a car, in mph, is described by the equation v(t)=-t^2+15. Describe the acceleration of the car at 5 seconds.
a. The car is slowing down at a rate of 10 mi./h^2.
b. The car is speeding up at a rate of 10 mi./h^2.
c. The car is slowing down at a rate of 12 mi./h^2.
d. The car is speeding up at a rate of 10 mi./h^2.
1.C
2.D
3.B
4.A
1. Well, I'd say let's calculate it together! The charge q(t) is given as 3t^2 + 2t - 5. To find the current at t = 0, we need to find the derivative of q(t) and evaluate it at t = 0. Taking the derivative, we get 6t + 2. Substituting t = 0, we get 6(0) + 2 = 2. So, the current at t = 0 is 2. Option c, you're the electric highlight of the circuit!
2. To find the instantaneous velocity, we need to find the derivative of the position equation f(t). So, let's do some math and calculate together! The derivative of f(t) = 4t^2 - 2t is given by v(t) = 8t - 2. Look at that, option d! The correct equation for the instantaneous velocity is v(t) = 8t - 2. Velocity, you're moving in the right direction!
3. Average velocity, huh? Let's calculate! The position equation f(t) is given as 90t^2 + 2t. To find the average velocity between minutes 3 and 7, we need to find the change in distance divided by the change in time. So, let's plug in the values! At t = 3, f(t) = 90(3^2) + 2(3) = 828. And at t = 7, f(t) = 90(7^2) + 2(7) = 4,616. The change in distance is 4,616 - 828 = 3,788 miles. The change in time is 7 - 3 = 4 minutes. Dividing the change in distance by the change in time, we get 3,788 / 4 = 947 miles per minute. Oh, so close! Option b, you're the average!
4. Acceleration, huh? Let's put the pedal to the metal and calculate together! The velocity equation v(t) = -t^2 + 15. To find the acceleration at 5 seconds, we need to find the derivative of v(t) and evaluate it at t = 5. Taking the derivative, we get -2t. Substituting t = 5, we get -2(5) = -10. So, the acceleration at 5 seconds is -10 mph^2. Option a, you're slowing it down!
did you ever get the answers to this i really need them.
David is correct.
Yes, David's answers are correct.
1. To find the current at t=0, we need to take the derivative of the charge equation q(t)=3t^2+2t-5 with respect to time. The derivative of q(t) gives us the current equation, i(t). The derivative of a quadratic function 3t^2+2t-5 is found by differentiating term by term. The derivative of 3t^2 is 6t, the derivative of 2t is 2, and the derivative of -5 is 0. Therefore, the current equation i(t) is given by i(t) = 6t + 2.
To find the current at t=0, we substitute t=0 into the current equation i(t).
i(0) = 6(0) + 2
i(0) = 0 + 2
i(0) = 2
Therefore, the current at t=0 is 2. The answer is option c.
2. The instantaneous velocity of an object is given by the derivative of its position function with respect to time. The given position equation is f(t) = 4t^2 - 2t. To find the derivative of f(t), we differentiate term by term. The derivative of 4t^2 is 8t, and the derivative of -2t is -2. Therefore, the instantaneous velocity equation v(t) is given by v(t) = 8t - 2.
Therefore, the instantaneous velocity at any given time, t, is v(t) = 8t - 2. The answer is option d.
3. The average velocity of an object over a time interval is given by the change in position divided by the change in time. In this case, we are asked to find the average velocity between minutes 3 and 7.
The position equation given is f(t) = 90t^2 + 2t. To find the average velocity between minutes 3 and 7, we need to calculate the change in position and change in time.
At t = 3 minutes, the position is given by f(3) = 90(3)^2 + 2(3) = 810 + 6 = 816 miles.
At t = 7 minutes, the position is given by f(7) = 90(7)^2 + 2(7) = 4410 + 14 = 4424 miles.
The change in position is f(7) - f(3) = 4424 - 816 = 3608 miles.
The change in time is 7 - 3 = 4 minutes.
Therefore, the average velocity between minutes 3 and 7 is 3608 miles / 4 minutes = 902 mi./min. The answer is option b.
4. The acceleration of a car is given by the derivative of its velocity function with respect to time. The given velocity equation is v(t) = -t^2 + 15. To find the derivative of v(t), we differentiate term by term. The derivative of -t^2 is -2t, and the derivative of 15 is 0. Therefore, the acceleration equation a(t) is given by a(t) = -2t.
To find the acceleration at 5 seconds, we substitute t=5 into the acceleration equation a(t).
a(5) = -2(5)
a(5) = -10
Therefore, the acceleration of the car at 5 seconds is -10 mi./h^2. This means the car is slowing down at a rate of 10 mi./h^2. The answer is option a.
Hey, you try first. You are supposed to know how to take the derivative of a polynomial.
if y = a x^n
then dy/dx = n a x^(n-1)
and the derivative of a constant is therefore 0
because if
y = a x^0
dy/dx = a * 0 * x^-1 = 0
so if
v = -t^2+15
then
a = dv/dt = -2 t + 0
and at t = 5
a = -2(5) = -10
if dv/dt is negative and v is positive, it is slowing.