A speeder passes a parked police car at 30 m/s and continues on, obliviously maintaining a constant velocity. At that instant, the police car starts from rest with a uniform acceleration of 2.44 m/s2.

a) How much time passes before the officer catches up to the speeder?
b) How far does the police car travel in that time?
c) How fast is the police car going when it catches up with the speeder?

Please help! ASAP!!

To solve this problem, we can use the equations of motion for uniformly accelerated motion. Let's break the problem down step by step:

a) How much time passes before the officer catches up to the speeder?

To find the time it takes for the police car to catch up to the speeder, we need to apply the equation of motion:

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

In this equation:
s is the distance traveled by the police car
u is the initial velocity of the police car (0 m/s because it starts from rest)
a is the acceleration of the police car (2.44 m/s^2)
t is the time taken

We know that the speeder is initially 30 m ahead of the police car. So we need to set up an equation by equating the distances traveled by both the speeder and the police car:

30 m + (30 m/s)t = (1/2)(2.44 m/s^2)t^2

Simplifying the equation:

(1.22 m/s^2)t^2 + 30t - 30 = 0

Now we can solve this quadratic equation to find the value of t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 1.22 m/s^2, b = 30 m/s, and c = -30. Plug in these values and solve for t.

By solving this equation, we find that there are two solutions for t: t1 ≈ -1.91 s and t2 ≈ 13.91 s. We disregard the negative value as time cannot be negative.

Therefore, it takes approximately 13.91 seconds for the police car to catch up to the speeder.

b) How far does the police car travel in that time?

To find the distance traveled by the police car, we'll use the equation:

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

Here, u is the initial velocity of the police car (0 m/s), t is the time (13.91 s), and a is the acceleration of the police car (2.44 m/s^2).

Plugging in the values, we get:

s = 0 + (1/2)(2.44 m/s^2)(13.91 s)^2

Simplifying the equation, we find:

s ≈ 176.7 meters

Therefore, the police car travels approximately 176.7 meters in that time.

c) How fast is the police car going when it catches up with the speeder?

To find the velocity of the police car when it catches up with the speeder, we can use the equation:

v = u + at

Here, u is the initial velocity of the police car (0 m/s), a is the acceleration of the police car (2.44 m/s^2), and t is the time (13.91 s).

Plugging in the values, we get:

v = 0 + (2.44 m/s^2)(13.91 s)

Simplifying the equation, we find:

v ≈ 34 m/s

Therefore, the police car is traveling at approximately 34 m/s when it catches up with the speeder.

Please note that all calculations are approximations and may vary slightly depending on rounding.

To find the answers to these questions, we can use the equations of motion.

a) To find the time it takes for the police car to catch up to the speeder, we need to determine the time it takes for the police car to cover the same distance that the speeder has already covered. 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.

Substituting the given values:
Initial velocity (for the police car): u = 0 m/s
Acceleration (for the police car): a = 2.44 m/s^2
Final velocity (for the police car): v = 30 m/s

\[t = \frac{v - u}{a} = \frac{30 - 0}{2.44}\]

Calculating this expression, we find that t is approximately 12.30 seconds.

Therefore, it takes approximately 12.30 seconds for the police car to catch up to the speeder.

b) To find the distance that the police car travels in that time, we can use the equation:

\[s = ut + \frac{1}{2}at^2,\]

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Substituting the given values:
Initial velocity (for the police car): u = 0 m/s
Acceleration (for the police car): a = 2.44 m/s^2
Time (for the police car to catch up to the speeder): t = 12.30 s

\[s = 0 \times 12.30 + \frac{1}{2} \times 2.44 \times (12.30)^2\]

Calculating this expression, we find that s is approximately 182.97 meters.

Therefore, the police car travels approximately 182.97 meters in that time.

c) To find the final velocity of the police car when it catches up to the speeder, we 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.

Substituting the given values:
Initial velocity (for the police car): u = 0 m/s
Acceleration (for the police car): a = 2.44 m/s^2
Time (for the police car to catch up to the speeder): t = 12.30 s

\[v = 0 + 2.44 \times 12.30\]

Calculating this expression, we find that v is approximately 30 meters per second.

Therefore, the police car will be traveling at approximately 30 meters per second when it catches up to the speeder.

d = 30 t

d = (1/2)(2.44)t^2
so
30 = 1.22 t
t = 30/1.22
solve for t and d
then
v = 2.44 t