<span>(a) 12.02 m/s
(b) 52.2 meters
This problem is an example of integral calculus. You've been given an acceleration vector which is usually known as the 2nd derivative. From that you need to calculate the velocity function (1st derivative) and position (actual function) by successively calculating the anti-derivative. So:
A(t) = 6.30 - 2.20t
V(t) = 6.30t - 1.10t^2 + C
We now have a velocity function, but need to determine C. Since we've been given the velocity at t = 0, that's fairly trivial.
V(t) = 6.30t - 1.10t^2 + C
3 = 6.30*0 - 1.10*0^2 + C
3 = 0 + 0 + C
3 = C
So the entire velocity function is:
V(t) = 6.30t - 1.10t^2 + 3
V(t) = -1.10t^2 + 6.30t + 3
Now for the location function which is the anti-derivative of the velocity function.
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
Now we need to calculate C. And once again, we've been given the location for t = 0, so
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
7.3 = -0.366666667*0^3 + 3.15*0^2 + 3*0 + C
7.3 = 0 + 0 + 0 + C
7.3 = C
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
Now that we have the functions, they are:
A(t) = 6.30 - 2.20t
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
let's answer the questions.
(a) What is the maximum speed achieved by the cyclist?
This can only happen at those points that meet either of the following criteria.
1. The derivative is undefined for the point.
2. The value of the derivative is 0 for the point.
As it turns out, the 1st derivative of the velocity function is the acceleration function which we have. So
A(t) = 6.30 - 2.20t
0 = 6.30 - 2.20t
2.20t = 6.30
t = 2.863636364
So one of V(0), V(2.863636364), or V(6) will be the maximum value. Therefore:
V(0) = 3
V(2.863636364) = 12.0204545454545
V(6) = 1.2
So the maximum speed achieved is 12.02 m/s
(b) Total distance traveled?
L(0) = 7.3
L(6) = 59.5
Distance traveled = 59.5 m - 7.3 m = 52.2 meters</span>
Answer:
By induction method
Explanation:
Induction method involves charging an electrically neutral body by bringing it in contact with an electrically charged body.
For the electrophus, a charge opposite that on the slab is induced on the side in contact with the slab; driving the opposite charge (this will be the same as that on the slab) to the other end of the elctrophus. Touching the electrophus removes the charge opposite the charge induced on the electrophus by the charged slab either by drawing up charge from the earth or taking the charge to earth (depends on the charge. A negative charge is drawn to earth while a positive charge draws up electrons from the earth)
It would be 2 to 4 seconds (:
Answer:
also tripled, factor of 3
Explanation:
equation of momentum is p = m * v , where p=momentum, m=mass, and v=velocity
if m is unchanged
p1 = m1 *v1
p2 = m2 *v2
m1 = m2 and v2 = 3*v1
p2 = m1 * 3v1 = 3*p1, tripled momentum
The answer is , Conductance
