Two cars are traveling along a straight line in the same direction, the lead car at 24.5m/s and the other car at 29.8m/s. At the moment the cars are 41.4m apart, the lead driver applies the brakes, causing her car to have an acceleration of -2.05m/s2.
A) How long does it take for the lead car to stop?
B) What is the distance it travels during this time?
C)Assuming that the chasing car brakes at the same time as the lead car, what must be the chasing car's minimum negative acceleration so as not to hit the lead car?
DHow long does it take for the chasing car to stop?
I have found that it takes the lead car 11.95 s to stop and it traveled 146.4 m. But still need help with the rest.
A) To find the time it takes for the lead car to stop, we can use the equation of motion:
v = u + at
where v is the final velocity (0 m/s in this case), u is the initial velocity (24.5 m/s in this case), a is the acceleration (-2.05 m/s^2), and t is the time we need to find.
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we have:
t = (0 - 24.5) / -2.05
t ≈ 11.95 seconds
Therefore, it takes approximately 11.95 seconds for the lead car to stop.
B) The distance traveled by the lead car during this time can be calculated using the formula:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the values, we have:
s = 24.5 * 11.95 + (1/2) * (-2.05) * (11.95)^2
s ≈ 145.9 meters
Therefore, the lead car travels approximately 145.9 meters during this time.
C) To find the minimum negative acceleration of the chasing car so it does not hit the lead car, we need to consider the relative motion between the two cars.
Since both cars are slowing down, the chasing car needs to have a maximum deceleration that matches the lead car's deceleration to avoid a collision. Therefore, the chasing car's minimum negative acceleration will be equal in magnitude to the lead car's negative acceleration, but with a positive sign.
So, the minimum negative acceleration of the chasing car is 2.05 m/s^2.
D) To find the time it takes for the chasing car to stop, we can use the same equation of motion:
t = (v - u) / a
where v is the final velocity (0 m/s in this case), u is the initial velocity (29.8 m/s in this case), and a is the acceleration (-2.05 m/s^2).
Substituting the values, we have:
t = (0 - 29.8) / -2.05
t ≈ 14.54 seconds
Therefore, it takes approximately 14.54 seconds for the chasing car to stop.