Figure 1 shows a red car and a green car that move toward each other. Figure 2 is a graph of their motion, showing the positions xg0 = 270 m and xr0 = -35.0 m at time t = 0 (time they intersect =12s). The green car has a constant speed of 20.0 m/s and the red car begins from rest. What is the acceleration magnitude of the red car?
To find the acceleration magnitude of the red car, we can use the equation for position as a function of time:
x = x0 + v0t + (1/2)at^2
where:
x is the final position
x0 is the initial position
v0 is the initial velocity
t is the time
a is the acceleration
In this case, the red car starts from rest (v0 = 0) and the green car has a constant speed of 20.0 m/s. We are given that the positions of the cars at time t = 0 are xg0 = 270 m and xr0 = -35.0 m.
At the time they intersect, the positions of the cars are equal (xg = xr), so we can set up an equation:
xg = xr
Taking the equation for the green car:
xg = xg0 + vgt
Substituting the given values:
xg = 270 m + (20.0 m/s)(t)
Taking the equation for the red car:
xr = xr0 + vrt + (1/2)art^2
Substituting the given values:
xr = -35.0 m + (0)(t) + (1/2)a(t)^2
xr = -35.0 m + (1/2)a(t)^2
Since xg = xr, we can set up an equation:
270 m + (20.0 m/s)(t) = -35.0 m + (1/2)a(t)^2
Simplifying the equation:
305 m + (20.0 m/s)(t) = (1/2)a(t)^2
We know that the time when the cars intersect is t = 12 s, so we can plug in this value:
305 m + (20.0 m/s)(12 s) = (1/2)a(12 s)^2
Simplifying further:
305 m + 240 m = 6a(12 s)^2
545 m = 6a(12 s)^2
Finally, we can solve for the acceleration magnitude:
a = (545 m) / (6(12 s)^2)
a ≈ 0.378 m/s^2
Therefore, the magnitude of the acceleration of the red car is approximately 0.378 m/s^2.
To find the acceleration magnitude of the red car, we can use the kinematic equation that relates position, initial velocity, time, and acceleration:
x = xi + vit + (1/2)at^2,
where x is the final position, xi is the initial position, vi is the initial velocity, t is the time, and a is the acceleration.
In this case, the red car starts from rest, so its initial velocity vi = 0 m/s. The green car has a constant speed of 20.0 m/s, so its initial velocity vi = 20.0 m/s. The initial positions of the cars are xg0 = 270 m and xr0 = -35.0 m.
At the time they intersect, t = 12 s, the final positions of the cars are xg = xg0 + vgt = 270 + 20 * 12 = 510 m and xr = xr0 + vrt + (1/2)art^2.
Since the cars intersect, their final positions are the same, so we can set xr = xg and solve for the acceleration ar:
xr0 + vrt + (1/2)art^2 = xg0 + vgt,
-35 + ar * 12^2/2 = 270 + 20 * 12.
Simplifying the equation:
ar * 144/2 = 270 + 240,
72 * ar = 510,
ar = 510 / 72,
ar = 7.08 m/s^2.
Therefore, the acceleration magnitude of the red car is approximately 7.08 m/s^2.
Figure 4: a moving car
if they intersect at t=12 sec, the green car traveled 20*12 m=240m
so the distance the red car travled in 12 sec is 305-240 m or 65m
distancered=vi+at=a*12 but that equals 65 m
a= 65/12 m/s^2'check my math