A motorist drives along a straight road at a constant speed of 15.0 m/s. just as she passes a parked police officer,the officer starts to accelerate at 2.00 m/s^2 to overtake her. assuming the officer maintains this acceleration, a) determine the time it takes the police officer to reach the motorist. find b) the speed and c) the total displacement of the officer as he overtakes the motorist.
a. d1 = d2,
15m/s * t = 0.5at^2,
15t = 0.5 * 2t^2,
15t = t^2,
t^2 - 15t = 0,
t(t - 15) = 0,
t = 0,
t - 15 = 0,
t = 15s.
b. d = 0.5 * 2 *(15)^2 = 225m.
post.
To determine the time it takes for the police officer to reach the motorist, we can use the equation of motion:
s = ut + (1/2)at^2
where:
s = displacement
u = initial velocity
a = acceleration
t = time
For the motorist, since she is driving at a constant speed of 15.0 m/s, her acceleration is 0. Thus, the equation simplifies to:
s = ut
For the police officer, his initial velocity is 0 since he starts from rest as the motorist passes him. The acceleration of the police officer is 2.00 m/s^2. Let's assume that the time taken for the police officer to overtake the motorist is t1.
For the motorist:
s1 = 15.0t1
For the police officer:
s2 = (1/2)(2.00)t1^2
Now, since both the motorist and the police officer reach the same position when they overtake each other, we can equate s1 and s2:
15.0t1 = (1/2)(2.00)t1^2
To solve for t1, we can rearrange the equation and solve for t1:
0.5t1^2 - 15.0t1 = 0
0.5t1(t1 - 30.0) = 0
Using the zero product property, we know that either t1 = 0 or (t1 - 30.0) = 0. Since time cannot be negative, we discard t1 = 0 and only consider:
t1 - 30.0 = 0
t1 = 30.0 seconds
a) The time it takes the police officer to reach the motorist is 30.0 seconds.
To find the speed (b) and the total displacement (c) of the officer as he overtakes the motorist, we can use the equations:
b) speed = initial velocity + acceleration * time
c) displacement = (initial velocity * time) + (0.5 * acceleration * time^2)
For the police officer:
b) speed = 0 + 2.00 * 30.0
= 60.0 m/s
c) displacement = (0 * 30.0) + (0.5 * 2.00 * 30.0^2)
= 900.0 m
b) The speed of the police officer when he overtakes the motorist is 60.0 m/s.
c) The total displacement of the police officer as he overtakes the motorist is 900.0 m.
To solve this problem, we can use the equations of motion:
1. For the motorist:
- Initial velocity (u) = 15.0 m/s
- Acceleration (a) = 0 (since the motorist maintains a constant speed)
- Time (t) = ?
- Displacement (s) = ?
2. For the police officer:
- Initial velocity (u) = 0 (since the officer is parked)
- Acceleration (a) = 2.00 m/s^2
- Time (t) = ?
- Displacement (s) = ?
a) To find the time it takes the police officer to reach the motorist, we can equate their displacements: s(motorist) = s(officer).
For the motorist:
s(motorist) = u(motorist) * t(motorist)
Since the motorist maintains a constant speed, the displacement is simply s(motorist) = 15.0 m/s * t(motorist).
For the police officer:
We need to calculate the time it takes for the officer to reach the motorist, so we'll use the equation: s(officer) = ut(officer) + (1/2)at^2.
Since the initial velocity of the officer is 0, this simplifies to s(officer) = (1/2)at^2.
Equating their displacements:
15.0 m/s * t(motorist) = (1/2) * 2.00 m/s^2 * t(officer)^2.
Now, let's solve for t(officer):
15.0 m/s * t(motorist) = (1/2) * 2.00 m/s^2 * t(officer)^2.
Simplifying the equation:
t(officer)^2 = (15.0 m/s * t(motorist)) / (1.00 m/s^2).
t(officer)^2 = 15.0 * t(motorist).
From this equation, we can see that t(officer) is directly proportional to t(motorist). If t(motorist) = t, then t(officer) = k * t.
Substituting back into the equation: t(officer)^2 = 15.0 * t, we get:
(k * t)^2 = 15.0 * t
Expanding the equation: k^2 * t^2 = 15.0 * t
Rearranging the equation: k^2 * t = 15.0
Solving for k: k^2 = 15.0 / t
Taking the square root: k = sqrt(15.0 / t)
Now we need to determine the value of k. We can do this by considering the initial conditions: when t = 0, the officer is at rest, so s(officer) = 0. Therefore, the time required for the officer to reach the motorist should be t(officer) = 2t.
Substituting this into the equation: k = sqrt(15.0 / (2t))
Now we can calculate the time it takes for the officer to reach the motorist.
b) The speed of the police officer when he overtakes the motorist can be determined by using the equation: v = u + at.
Since the officer starts from rest, the initial velocity (u) is 0. Substituting the values into the equation:
v = 0 + 2.00 m/s^2 * t(officer).
c) The total displacement of the police officer as he overtakes the motorist can be calculated using the equation: s = ut + (1/2)at^2, where u is the initial velocity (0 m/s), a is the acceleration (2.00 m/s^2), and t is the time.
Now, let's solve these equations to find the answers.