Suppose that during a test drive of two cars, one car travels 224 miles in the same time that a second car travels 160 miles. I
1 answer:
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1) The law of motion of the projectile is
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
1) Maximum
2) Maximum
Explanation:
The force acting on a mass on a spring is given by Hooke's law; in magnitude:
where
F is the force
k is the spring constant
x is the displacement
Also we know from Newton's second law that we can write
where
m is the mass
a is the acceleration
So we can write the equation as
(1)
From this relationship, we see that the acceleration is directly proportional to the displacement.
On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by
(2)
where the first term is the elastic potential energy while the second term is the kinetic energy, and where
v is the velocity of the mass
From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.
Therefore this means that:
- When acceleration increases, velocity decreases
- When acceleration decreases, velocity increases
Therefore, the two answers here are:
- When the acceleration of a mass on a spring is zero, the velocity is at a maximum
When the velocity of a mass on a spring is zero, the acceleration is at a maximum