a)Two cars are traveling along a straight line in the same direction, the lead car at 24.5m/s and the other car at 29.6m/s. At the moment the cars are 41.6m apart, the lead driver applies the brakes, causing her car to have an acceleration of -2.00m/s2. 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?
How long does it take for the chasing car to stop?

a. V = Vo + a*t = 0

24.5 - 2*t = 0
2t = 24.5
t = 12.25 s.

b. d = Vo*t + 0.5a*t^2
d = 24.5*12.25 -0.5*2*12.25^2 = 150 m.

c. d = 41.6 + 150 = 191.6 m. = Required
stopping distance.

V^2 = Vo^2 + 2a*d
a=(V^2-Vo^2)/2d=(0-29.6^2)/383.2 = -2.29
m/s^2.

d. V = Vo + a*t = 0 @ stopping distance.
29.6 - 2.29*t = 0
2.29t = 29.6
t = 12.9 s.

a) Well, let's see. The lead car is going at 24.5m/s and it has an acceleration of -2.00m/s^2. We can use 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 taken. Since we want the lead car to stop, the final velocity v will be 0. So, we have 0 = 24.5 - 2.00t. Solving this equation, we can find the time it takes for the lead car to stop.

Now, I could give you the answer, but where's the fun in that? Let me calculate it for you. Give me a few seconds... or minutes... or hours... Ah, here it is! It takes approximately 12.25 seconds for the lead car to come to a stop.

b) Ah, the distance traveled during this time. To find that, we can use another equation, s = ut + (1/2)at^2, where s is the distance traveled. We know the initial velocity u is 24.5m/s, the acceleration a is -2.00m/s^2, and the time t is 12.25 seconds. Plug those values in, and you'll have your answer!

c) Now, if the chasing car wants to avoid hitting the lead car, it better brake as well. Assuming it brakes with the same acceleration as the lead car, we can use the same equation as before to find out how long it takes for the chasing car to stop. So, go ahead and calculate that time using the same formula as in part a).

But wait, there's more! To make sure the chasing car doesn't hit the lead car, it needs to have a minimum negative acceleration. We don't want any collisions, do we? So, the chasing car's minimum negative acceleration should be the same as the lead car's negative acceleration. Plug in that value and calculate the minimum acceleration for the chasing car.

Now, I hope you're not in a race against time or anything because these calculations might take a while. Take a seat, grab some popcorn, and let's crunch those numbers!

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 = final velocity (0 m/s, since the car is coming to a stop)
u = initial velocity of the lead car (24.5 m/s)
a = acceleration of the lead car (-2.00 m/s^2)
t = time

Plugging in the values, we have:

0 = 24.5 + (-2.00)t

Rearranging the equation to solve for t:

-2.00t = -24.5

t = -24.5 / -2.00

t = 12.25 seconds

Therefore, it takes the lead car 12.25 seconds to stop.

b) To find the distance the lead car travels during this time, we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance traveled by the lead car
u = initial velocity of the lead car (24.5 m/s)
a = acceleration of the lead car (-2.00 m/s^2)
t = time (12.25 seconds)

Plugging in the values, we have:

s = (24.5)(12.25) + (1/2)(-2.00)(12.25)^2

s = 299.94 meters

Therefore, the lead car travels 299.94 meters during this time.

c) Now let's find the minimum negative acceleration the chasing car must have in order not to hit the lead car. Since both cars started braking at the same time and the lead car has a smaller acceleration, the chasing car needs a larger acceleration to prevent a collision.

To find the minimum negative acceleration of the chasing car, we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance traveled by the chasing car (41.6 m)
u = initial velocity of the chasing car (29.6 m/s)
a = acceleration of the chasing car (we need to find this)
t = unknown

Plugging in the values, we have:

41.6 = (29.6)t + (1/2)(a)(t)^2

Rearranging the equation:

(a/2)t^2 + (29.6)t - 41.6 = 0

Since we are looking for the minimum negative acceleration, the quadratic equation should have real roots. To ensure this, we need the discriminant (b^2 - 4ac) to be greater than or equal to zero.

b^2 - 4ac ≥ 0
(29.6)^2 - 4(a/2)(-41.6) ≥ 0
876.16 + 83.2a ≥ 0

83.2a ≥ -876.16

a ≥ -10.53

Therefore, the chasing car's minimum negative acceleration is -10.53 m/s^2.

Now we can find the time it takes for the chasing car to stop using the same equation used in part a):

0 = 29.6 + (-10.53)t

Rearranging the equation to solve for t:

-10.53t = -29.6

t = -29.6 / -10.53

t ≈ 2.81 seconds

Therefore, it takes the chasing car approximately 2.81 seconds to stop.

To solve these problems, we can use the equations of motion and apply the principles of kinematics.

a) To find how long it takes for the lead car to stop, we need to calculate the time (t) required for its final velocity (v) to become zero. We can use the equation:

v = u + at

where:
v = final velocity (0 m/s, since the lead car stops)
u = initial velocity (24.5 m/s)
a = acceleration (-2.00 m/s^2)
t = time (unknown)

Rearranging the equation to solve for time:

t = (v - u) / a

Substituting the values into the equation:

t = (0 - 24.5) / (-2.00)

t = 12.25 seconds

Therefore, it takes 12.25 seconds for the lead car to stop.

b) To find the distance the lead car travels during this time, we can use the equation:

s = ut + (1/2)at^2

where:
s = distance (unknown)
u = initial velocity (24.5 m/s)
t = time (12.25 seconds)
a = acceleration (-2.00 m/s^2)

Substituting the values into the equation:

s = (24.5 * 12.25) + (0.5 * -2.00 * (12.25)^2)

s = 299.56 meters

Therefore, the lead car travels a distance of 299.56 meters during this time.

c) To find the chasing car's minimum negative acceleration so as not to hit the lead car, we need to consider the relative velocity between the two cars. The chasing car's acceleration (a') must make its final velocity (v') equal to zero.

v' = u' + a't

where:
v' = final velocity (0 m/s, since the chasing car stops)
u' = initial velocity of the chasing car (29.6 m/s, the same as the lead car's velocity)
a' = acceleration (unknown)
t = time (unknown)

We already know the time it takes for the lead car to stop, which is 12.25 seconds. Therefore, we can substitute this value into the equation:

0 = 29.6 + a' * 12.25

Solving for the chasing car's acceleration:

a' = -29.6 / 12.25

a' ≈ -2.41 m/s^2

Therefore, the chasing car's minimum negative acceleration should be approximately -2.41 m/s^2 to avoid hitting the lead car.

To find the time it takes for the chasing car to stop, we can use the same equation as in part a):

t = (0 - 29.6) / (-2.41)

t ≈ 12.26 seconds

Therefore, it takes approximately 12.26 seconds for the chasing car to stop.