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
Answer:
B. A object in motion stays in motion, and an object at rest stays at rest unless acted upon by a net force.
Answer:
the formula for calculating acceleration is ending speed minus starting speed divided by time.
Considering that while traveling on a road with a<u> final speed of 15 m/s</u>, and an<u> initial speed of 24 m/s</u>, with a given time <u>of 12 seconds.</u>
To calculate the acceleration, we apply the following formula:
α = Vf - Vo/t
We add our data into the formula and solve:
α = 15 m/s - 24 m/s/12 sec
α = -0.75 m/s²
Therefore, the acceleration of the car is -0.75 m/s².
<h2>Skandar</h2>
12 newtons is your answer
