The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal
1 answer:
You might be interested in
Answer: Hello there!
We know this:
The distance between the cars at t= 0 is D.
car 2 has an initial velocity of v0 and no acceleration.
car 1 has no initial velocity and a acceleration of ax that starts at t = 0
then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.
Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:
position of car 1 + position of car 2 = D
and in this way we could ignore constants of integration :D
for the position of each car we integrate again:
P1(t) = (1/2)ax*t^2 and P2(t) = v0t
v0t + (1/2)ax*t^2 = D
v0t + (1/2)ax*t^2 - D = 0
now we can solve it for t using the Bhaskara equation.
that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).
Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:
where again, you need to replace the values of v0, D and ax.
- The largest increase in potential energy occurs when the charge is moving north. This is because the charge is negative, so it acquires potential energy when moving in the same direction of the field (viceversa, a positive charge when moving in the direction of the field it loses potential energy converting it into kinetic energy). The amount of potential energy gained is equal to the product of the charge and the distance covered:
- The second largest increase is when the charge is moving east. In this case, actually, the variation of potential energy is zero. This is because the charge is moving perpendicular to the field, and so it is moving along points with same potential. Therefore, in this case the variation of potential energy is zero:
- Finally, when the charge is moving south, it loses potential energy. This is because it is moving against the electric field, and since it is a negative charge, in this direction it loses potential energy converting it into kinetic energy. Therefore, in this case:
- The second largest increase is when the charge is moving east. In this case, actually, the variation of potential energy is zero. This is because the charge is moving perpendicular to the field, and so it is moving along points with same potential. Therefore, in this case the variation of potential energy is zero:
- Finally, when the charge is moving south, it loses potential energy. This is because it is moving against the electric field, and since it is a negative charge, in this direction it loses potential energy converting it into kinetic energy. Therefore, in this case: