A) Car A is initially ahead
B) The two cars are at the same point at the times: t = 0, t = 2.27 s and
t = 5.73 s
C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s
D) The two cars have same acceleration at t = 2.67 s
Explanation:
A)
The position of the two cars at time t is given by the following functions:
with
Substituting,
And
with
Substituting,
Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:
So, car A is initially ahead.
B)
The two cars are at the same point when their position is the same. Therefore, when
which means when
Re-arranging the equation, we find
One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation
which has two solutions:
t = 2.27 s
t = 5.73 s
So, these are the times at which the cars are at the same point.
C)
The distance between the two cars A and B is not changing when the velocities of the two cars is the same.
The velocity of car A is given by the derivative of the position of car A:
The velocity of car B is given by the derivative of the position of car B:
Therefore, the distance between the two cars is not changing when the two velocities are equal:
This is another second-order equation, which has two solutions:
t = 1.00 s
t = 4.33 s
D)
The acceleration of each car is given by the derivative of the velocity of the car A.
The acceleration of car A is:
While the acceleration of car B is:
So, the two cars have same acceleration when
And solving the equation, we find:
So, the two cars have same acceleration at t = 2.67 s.
Learn more about accelerated motion:
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