a drunken motorist who is moving at a constant velocity of 90 km/h passes a stationary police patrol car. The patrol car immediately gives chase at a constant acceleration and catches up with the motorist after a distance of 10 km. Calculate the time taken by the patrol car to catch up with the motorist.
Start counting time t from when the drunk motorist passes. Let his distance travelled be X1 and the distance travelled by the police car be X2.
X1 = V*t
X2 = (a/2) t^2
X1 = X2 when both equal 10 km. That is when the police car catches up.
t = 10 km/(90 km/h) = 1/9 h = 6.67 minutes
You could also solve for the police car's acceleration a, but they don't ask for that.
Well, I'm not sure I should be encouraging drunken driving, but I can help with the math part. So, let's see: the drunken motorist is moving at 90 km/h and the patrol car catches up after a distance of 10 km. To figure out the time taken, we need to find the acceleration of the patrol car.
Since the motorist is moving at a constant velocity, we know that their acceleration is 0. So, the acceleration of the patrol car is what allows it to catch up.
Now, we can use the equation of motion:
Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2
We have the distance (10 km) and the initial velocity (0 km/h for the patrol car), so we can solve for time.
10 = 0 * t + (1/2) * a * t^2
Since the motorist is moving at a constant velocity, we know the patrol car's acceleration is equal to the acceleration of the motorist. So, we can write:
10 = (1/2) * a * t^2
Now, we need to solve for t. Hmm, this is turning into a serious math problem. Maybe I should've stuck to jokes... Oh well, let's brave through it!
First, let's multiply both sides of the equation by 2 to get rid of the (1/2):
20 = a * t^2
Then, divide both sides by a:
20/a = t^2
And finally, take the square root of both sides:
√(20/a) = t
So, there you have it! The time taken by the patrol car to catch up with the motorist is √(20/a). Just try to remember, it's better to have sober drivers on the road and leave the chasing to the professionals. Stay safe out there!
To solve this problem, we can use the equation of motion for the patrol car:
\[d = v_0t + \frac{1}{2}at^2\]
Where:
- d is the distance traveled by the patrol car (10 km)
- \(v_0\) is the initial velocity of the patrol car (0 km/h since it was stationary initially)
- a is the constant acceleration of the patrol car (which we need to find)
- t is the time taken by the patrol car to catch up with the motorist (which we need to find)
Now, let's find the acceleration of the patrol car.
Since the drunken motorist is traveling at a constant velocity, we know that his velocity is \(v_1 = 90\) km/h, and since the patrol car catches up to him, their velocities become the same at that point.
Let's denote the time it takes for the patrol car to catch up with the motorist as \(t_1\). At this point, the velocities of both vehicles are equal, so we can write the equation:
\[v_0 + at_1 = v_1\]
Since the initial velocity of the patrol car is 0 km/h, the equation becomes:
\[0 + at_1 = 90\]
Now we can solve for the acceleration:
\[a = \frac{v_1}{t_1} = \frac{90}{t_1}\]
Now, let's substitute this acceleration value into the equation of motion for the patrol car:
\[10 = 0 + \frac{1}{2} \cdot \frac{90}{t_1} \cdot t_1^2\]
Simplifying the equation gives:
\[10 = \frac{45}{t_1} \cdot t_1^2\]
Simplifying further:
\[10 = 45t_1\]
Finally, we can solve for \(t_1\):
\[t_1 = \frac{10}{45}\]
Therefore, the time taken by the patrol car to catch up with the motorist is \(t_1 = \frac{10}{45}\) hours.
To calculate the time taken by the patrol car to catch up with the motorist, we can use the concepts of relative velocity and equations of motion.
Step 1: Determine the relative velocity between the patrol car and the motorist.
The relative velocity is the difference between the velocities of the patrol car and the motorist. In this case, the motorist is moving at a constant velocity of 90 km/h, and the patrol car starts from rest and accelerates to catch up with the motorist.
Step 2: Convert the velocities to meters per second (m/s).
To ensure consistency in units, convert the velocities to meters per second. 1 km/h is equal to (5/18) m/s.
So, the motorist's velocity is v1 = 90 km/h * (5/18) m/s = 25 m/s.
Step 3: Use the equations of motion to determine the time taken to catch up.
We can use the equation: S = ut + (1/2)at^2, where:
S = distance traveled (10 km = 10000 m),
u = initial velocity of the patrol car (0 m/s),
a = acceleration of the patrol car, which is constant,
t = time taken to catch up.
Since the motorist is moving at a constant velocity, their displacement is equal to the distance traveled by the patrol car, which is 10 km.
So, 10000 m = (1/2) a t^2
From here, we have to find the acceleration.
Using another equation of motion, v = u + at, where:
v = final velocity of the patrol car (equal to the motorist's velocity, 25 m/s),
u = initial velocity of the patrol car,
a = acceleration of the patrol car,
t = time taken to catch up.
Rearranging the equation, we have:
a = (v - u) / t
Plugging in the values, (v = 25 m/s, u = 0 m/s), we get:
a = (25 - 0) / t
Combining the equations, we can solve for t:
10000 m = (1/2)((25 - 0) / t) t^2
Simplifying further, we have:
20000 = 25t
Now, divide both sides by 25:
t = 20000 / 25
Finally, calculate the value:
t = 800 seconds
Therefore, the patrol car takes 800 seconds to catch up with the motorist.