Two cars both start from rest and have the same constant acceleration. When radar is used to measure the velocity of each car, car A is found to be moving twice as fast as car B. At the instant the radar is used, car A has a displacement that is ______ times as great as the displacement of car B.
four
To solve this problem, let's consider the following:
Let's assume the initial velocity and acceleration of both cars are the same.
Let's denote:
- a as the acceleration of both cars,
- vA as the velocity of car A,
- vB as the velocity of car B, and
- dA as the displacement of car A,
- dB as the displacement of car B.
Given that car A is moving twice as fast as car B, we can write:
vA = 2vB ----(1)
We know that velocity is the derivative of displacement with respect to time. So, we can integrate the velocity equation (1) with respect to time to find the displacement equations for both cars.
∫vA dt = ∫2vB dt
dA = 2dB ----(2)
Therefore, at the instant when the radar is used:
The displacement of car A is twice as great as the displacement of car B.
In summary, the answer is 2. The displacement of car A is 2 times the displacement of car B.
To solve this problem, we need to use the equations of motion for uniformly accelerated motion. Let's assume that the initial velocity of both cars is zero (as stated in the question) and the acceleration is the same for both cars.
We can start by using the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
For car A, we have v_A = 2v_B (car A is moving twice as fast as car B). Since both cars start from rest, their initial velocities are zero. Therefore, we have:
v_A = 2v_B = 0 + a*t_A,
v_B = 0 + a*t_B,
where t_A and t_B are the times taken for cars A and B, respectively.
Rearranging these equations gives:
t_A = (2/v_B)*t_B,
Now, we need to use another equation to relate displacement, velocity, acceleration, and time. The equation is:
s = ut + (1/2)at²,
where s is the displacement.
For car A, the displacement is given by:
s_A = 0*t_A + (1/2)*a*t_A²,
s_A = (1/2)*a*(2/v_B)*(t_B)².
For car B, the displacement is:
s_B = 0*t_B + (1/2)*a*t_B²,
s_B = (1/2)*a*t_B².
Dividing the two equations, we get:
s_A/s_B = [(1/2)*a*(2/v_B)*(t_B)²]/[(1/2)*a*t_B²],
s_A/s_B = 2/v_B.
Therefore, at the instant the radar is used, car A has a displacement that is 2/v_B times greater than the displacement of car B.