Figure (a) shows a red car and a green car that move toward each other. Figure (b) is a graph of their motion, showing the positions xg0 = 270 m and xr0 = -35 m at time t = 0. The green car has a constant speed of 24 m/s and the red car begins from rest. What is the acceleration magnitude of the red car?
On figure b, a position graph. Where the meet, note the time, and position.
so the distance traveled by carr, is from -35 to the meeting position. So you have time in motion, and distance for car red.
d=vi*t+1/2 a t^2 vi=0, so you can figure a from your distance (from -35 to final position) and time.
To find the acceleration magnitude of the red car, we need to determine its change in velocity over time. We can do this by using the equation of motion:
Δv = v_f - v_i,
where Δv is the change in velocity, v_f is the final velocity, and v_i is the initial velocity.
In this case, the green car has a constant speed of 24 m/s, so its velocity remains unchanged throughout.
The initial velocity of the red car (v_i) is 0 m/s since it starts from rest. We need to find the final velocity of the red car (v_f).
From the given information in Figure (b), we can see that the positions of the red and green cars at time t = 0 are xg0 = 270 m and xr0 = -35 m, respectively.
Using the formula for calculating velocity given initial position (x0) and final position (x_f):
v = (x_f - x_0) / t,
we can calculate the final velocity of the red car:
v_f = (xr - xr0) / t,
where xr is the position of the red car at a later time.
Let's assume xr is the new position of the red car at time t.
We know from the problem statement that the green car moves toward the red car, so their positions will converge. This means that at some point, xr and xg will be the same, so:
xr = xg.
Therefore, we can substitute the positions:
xr = x_f = xg.
Now we can calculate the final velocity of the red car:
v_f = (xg - xr0) / t.
Since the green car travels at a constant speed of 24 m/s, the time it takes for the green car to cover the distance between the initial positions of the two cars is:
t = (xg - xr0) / v_g,
where v_g is the velocity of the green car.
Substituting the values:
t = (270 m - (-35 m)) / 24 m/s.
Simplifying:
t = 11.25 s.
Now that we have the time it takes for the green car to reach the initial position of the red car, we can calculate the final velocity of the red car using the equation we derived earlier:
v_f = (xg - xr0) / t,
v_f = (270 m - (-35 m)) / 11.25 s.
Simplifying:
v_f = 28 m/s.
Next, we can calculate the acceleration of the red car using the formula:
a = (v_f - v_i) / t,
where a is the acceleration we are looking for, v_f is the final velocity of the red car, v_i is its initial velocity (which is 0 m/s since it starts from rest), and t is the time it takes for the green car to reach the initial position of the red car.
Plugging in the values:
a = (28 m/s - 0 m/s) / 11.25 s.
Simplifying:
a = 2.49 m/s^2.
Therefore, the acceleration magnitude of the red car is approximately 2.49 m/s^2.