1.A motorist drive along a straight road at a constant speed of 15.0 m/s.Just as she passes a parked motorcycle police officer,the officer starts to accelerate at 2.0 m/s^2 to overtake her. Assuming the officer maintains this acceleration, a) determine the time it takes the police officer the motorist.
b)the speed,
c)the total displacement of the officer as ge overtakes the motorist.
2.A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of cliff 30.0 m high. At the same instant,another stone is thrown up with a speed of 40.0 m/s frim the foot of the cliff. Will the two stones meet before they hit the ground?give exact details.
thanks :)
See previous post: Tue, 9-9-14, 9:45 AM.
a) To determine the time it takes for the police officer to overtake the motorist, we need to use the formula:
\(s = ut + 0.5at^2\),
where:
- \(s\) is the displacement,
- \(u\) is the initial velocity,
- \(t\) is the time, and
- \(a\) is the acceleration.
Here, the motorist is moving at a constant speed, so her initial velocity \(u = 15.0 \, \text{m/s}\).
Since the police officer starts from rest (\(u = 0\)) and has an acceleration of \(2.0 \, \text{m/s}^2\), we can use the same formula:
\(s = 0.5at^2\).
Since the police officer needs to reach the same displacement as the motorist (to overtake her), we set \(s_{\text{motorist}} = s_{\text{officer}}\).
So, \(0.5 \times 2.0t^2 = 15.0t\).
Simplifying, we get \(t^2 = \frac{15.0}{2.0}\).
Taking the square root of both sides, we find \(t \approx 3.5 \, \text{s}\).
Therefore, it takes the police officer approximately 3.5 seconds to overtake the motorist.
b) To determine the speed of the police officer, we use the formula:
\(v = u + at\),
where:
- \(v\) is the final velocity.
Since the police officer has an initial velocity of \(0 \, \text{m/s}\) and an acceleration of \(2.0 \, \text{m/s}^2\), we can substitute these values into the formula:
\(v = 0 + 2.0 \times 3.5\).
Simplifying, we find \(v \approx 7.0 \, \text{m/s}\).
Therefore, the speed of the police officer is approximately \(7.0 \, \text{m/s}\).
c) To determine the total displacement of the police officer as they overtake the motorist, we can use the formula:
\(s = ut + 0.5at^2\).
In this case, since the initial velocity is \(0 \, \text{m/s}\) and the time is \(3.5 \, \text{s}\), we can substitute these values into the formula:
\(s = 0 + 0.5 \times 2.0 \times (3.5)^2\).
Simplifying, we find \(s \approx 12.3 \, \text{m}\).
Therefore, the total displacement of the police officer as they overtake the motorist is approximately \(12.3 \, \text{m}\).
2. In the second question, we are given two stones being thrown vertically. The stone thrown from the edge of a cliff and the stone thrown from the foot of the cliff.
To determine if the two stones will meet before hitting the ground, we need to compare their respective heights and initial velocities.
The stone thrown from the edge of the cliff:
- Height: 30.0 m
- Initial velocity: 15.0 m/s (upward)
The stone thrown from the foot of the cliff:
- Height: 0.0 m (since it starts from the ground)
- Initial velocity: 40.0 m/s (upward)
To find out if the two stones will meet, we need to calculate the time it takes for each stone to reach the ground.
For the stone thrown from the edge of the cliff (let's call it Stone A), we can use the formula:
\(s = ut + 0.5gt^2\),
where:
- \(s\) is the displacement,
- \(u\) is the initial velocity,
- \(t\) is the time, and
- \(g\) is the acceleration due to gravity (-9.8 m/s², since it acts in the opposite direction of the upward motion).
Substituting the values for Stone A, we have:
\(30.0 = 15.0t + 0.5(-9.8)t^2\).
Simplifying, we get:
\(4.9t^2 + 15.0t - 30.0 = 0\).
Using the quadratic formula, we find two solutions: \(t \approx 1.527 \, \text{s}\) and \(t \approx -3.527 \, \text{s}\).
Since time cannot be negative, we consider only the positive solution as the time it takes for Stone A to reach the ground: \(t \approx 1.527 \, \text{s}\).
For the stone thrown from the foot of the cliff (let's call it Stone B), we can again use the same formula:
\(s = ut + 0.5gt^2\).
Substituting the values for Stone B, we have:
\(0.0 = 40.0t + 0.5(-9.8)t^2\).
Simplifying, we get:
\(4.9t^2 + 40.0t = 0\).
Factoring out \(t\), we have:
\(t(4.9t + 40.0) = 0\).
This equation has two solutions: \(t = 0\) (which is the initial time, not our interest) and \(t \approx -8.163 \, \text{s}\).
Again, since time cannot be negative, we only consider the positive solution: \(t \approx 8.163 \, \text{s}\).
Therefore, Stone B takes approximately 8.163 seconds to reach the ground.
Comparing the times, we see that Stone A takes 1.527 seconds, while Stone B takes 8.163 seconds. Therefore, Stone B will hit the ground before Stone A.
Hence, the two stones will not meet in the air.