Answer:
Although this question is not complete, I would give a general solution to this kind of problems.
If y(t) describes the position of a body with time such that
y(t) = at^(n) + bt^(m) + C
Then
V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)
Explanation:
As an example supplies the position of a particle is given by
y(t) = 4t³- 3t² + 9
V(t) = 4x3t²- 3x2t¹
V(t) = d(t)/dt = 12t² - 6t.
Another example,
If y(t) = 15t³ - 2t² + 30t -80
V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.
Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.
One last example,
y(t) = 2t^6 -3t²
V(t) = d(t)/dt = 12t^5 - 6t
I don’t think I need anything from my dad to get my hair fixed so I’m just bored I don’t want you guys
Answer:
Electric force results form mutual interactions between two charges.
Answer: The correct answer is option b.
Explanation: We are given that the rocket is at rest initially final velocity is 445m/s.
The acceleration of the rocket is 
To calculate the distance of rocket, we use third equation of motion, which is:
where, v = final velocity = 445m/s
u = initial velocity = 0m/s
a = acceleration = 
s = distance = ? m
Putting values in above equation, we get:
Answer:
1km = o.621371 mile
Explanation:
1.609 kilometers equal 1 mile. The kilometer is a unit of measurement, as is the mille. However, a mile is longer than a kilometer.
