(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.
Recall that
We have , so that
(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.
(c) The ball's average velocity is 0. Average velocity is given by , and we know that .
(d) The position of the ball at time is given by
Take the starting position to be the origin, . Then after 6 seconds,
so the ball is 42 m away from where it started.
We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:
Since the velocity is positive, the ball is still moving up the incline.
Theories
If it’s wrong oh well because I just guest buh glad to help :)
Time = 15/-0.5 = 14.5/-0.5 = 14.5s
The answer is 14.5s
Sorry no one would answer it sooner.
Answer:
Acceleration of the boy a₁:
Explanation:
Conceptual analysis
We apply Newton's second law to the boy and the girl:
F = m*a (Formula 1)
F : Force in Newtons (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Nomenclature
m₁ : boy mass
m₂ : girl mass
a₁ : boy acceleration
a₂ : girl acceleration
F₁ : boy acceleration
F₂ : girl acceleration
Known data
m₁ = 57 kg
m₂ = 41 kg
a₂ = 2.6 m/s²
Problem development
We apply to Newton's third law of action and reaction, then:
F₁ = F₂ , We apply the formula (1):
m₁*a₁ = m₂*a₂