a motorcycle traffic police starts from rest at point A two seconds after a car speeding at the constant rate of 120 km/h passes point A (where speed limit is 100 km/h only) if the policeman accelerates at the rate of 6 m/s^2 until he reaches the speed of 150 km/h and maintains constant speed thereafter. calculate the distance, s, from point A to the point at which he overtakes the speeding car
120km/hr = 33.333 m/s
so, at time t, the car has gone 33.333 t meters
how long does the cop take to achieve 150km/hr? That's 41.666 m/s, so at 6m/s^2 it takes 6.9444 seconds.
During the 6.9444 seconds, he travels 1/2 (6)(6.9444)^2 = 144.67 m
So, the cop overtakes the car when
33.333t = 144.67 + 41.666(t-2-6.9444)
t = 27.362
check:
car(27.362) = 33.333*27.362 = 912m
cop(25.362) = 144.47+41.666*18.418 = 912m
so, the cop overtakes the car 912m from A.
Well, well, well, let's help our motorcycle traffic police catch that speedy car!
First, let's convert the speeds to meters per second because we'll be using some physics formulas later on. So, 120 km/h is about 33.33 m/s, and 100 km/h is about 27.78 m/s. The police accelerates to 150 km/h, which is about 41.67 m/s.
Since the police starts from rest 2 seconds after the car passes point A, we need to find out how far the car traveled during those 2 seconds. The formula to calculate distance is speed multiplied by time. So, the car traveled 33.33 m/s multiplied by 2 seconds, which gives us 66.67 meters.
Now, let's focus on the police. The acceleration of the police is 6 m/s². We can use the formula:
distance = (final speed)² - (initial speed)² / (2 * acceleration)
Let's plug in the values! Final speed is 41.67 m/s, initial speed is 0 (as the police starts from rest), and the acceleration is 6 m/s². Solving this equation will give us the distance traveled by the police.
After finding the distance traveled by the police, we need to add the 66.67 meters the car traveled during those 2 seconds. That will give us the total distance, s, from point A to the point at which the police overtakes the car.
I hope you're ready for the final calculation because I'm getting all revved up!
To solve this problem, we need to find the time it takes for the motorcycle to catch up with the car and then calculate the distance traveled by the motorcycle.
Let's break down the problem step-by-step:
Step 1: Convert the speeds to meters per second (m/s) for easier calculations.
Car's speed = 120 km/h = (120 * 1000) / (60 * 60) = 33.33 m/s
Police motorcycle's speed limit = 100 km/h = (100 * 1000) / (60 * 60) = 27.77 m/s
Police motorcycle's maximum speed = 150 km/h = (150 * 1000) / (60 * 60) = 41.66 m/s
Step 2: Calculate the time it takes for the motorcycle to catch up with the car.
Let t be the time it takes for the motorcycle to catch up.
During this time, the car has been traveling for (t + 2) seconds.
Step 3: Calculate the distance traveled by the car during the time it took for the motorcycle to catch up.
Distance traveled by the car = car's speed × time = 33.33 m/s × (t + 2) s
Step 4: Calculate the distance traveled by the motorcycle during the time it took to catch up.
The motorcycle starts from rest, so we can use the equation of motion: s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
From rest, the initial velocity of the motorcycle is 0 m/s.
The motorcycle accelerates at a rate of 6 m/s^2 for t seconds until it reaches its maximum speed.
Distance traveled during acceleration = (1/2) × 6 m/s^2 × t^2
The motorcycle then travels at a constant speed of 41.66 m/s for (t + 2) seconds.
Distance traveled during constant speed = 41.66 m/s × (t + 2) s
Step 5: Set the distances traveled by the car and the motorcycle equal to each other and solve for t.
33.33 m/s × (t + 2) s = (1/2) × 6 m/s^2 × t^2 + 41.66 m/s × (t + 2) s
Step 6: Simplify and solve the quadratic equation.
33.33t + 66.66 = 3t^2 + 83.32t + 166.64
3t^2 + 49.99t - 100 = 0
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a = 3, b = 49.99, c = -100
Step 7: Calculate the positive value of t.
t = (-49.99 ± √(49.99^2 - 4 * 3 * -100)) / (2 * 3)
t ≈ 4.928 seconds
Since we cannot have negative time, we discard the negative solution.
Step 8: Calculate the distance traveled by the motorcycle.
Distance traveled by the motorcycle = distance during acceleration + distance during constant speed
= (1/2) × 6 m/s^2 × (4.928 s)^2 + 41.66 m/s × (4.928 s + 2 s)
≈ 0.5 × 6 m/s^2 × 24.282 s^2 + 41.66 m/s × 6.928 s
≈ 73.1 m + 288.67 m
≈ 361.77 m
Therefore, the distance from point A to the point at which the motorcycle overtakes the car is approximately 361.77 meters.
To calculate the distance from point A to the point where the motorcycle overtakes the car, we can break down the problem into different stages.
Stage 1: Determining time taken by the car to reach point A
The car is traveling at a constant rate of 120 km/h. We want to determine the time it takes for the car to reach point A. We know that speed = distance/time, so rearranging the formula, we have time = distance/speed. Given that the speed of the car is 120 km/h and the distance is from point A to the starting point, we can calculate the time it takes for the car to reach point A.
Stage 2: Determining the time taken by the motorcycle to catch up with the car
We know that the motorcycle starts from rest at point A, two seconds after the car passes point A. Therefore, we need to calculate the time it takes for the motorcycle to catch up with the car after the motorcycle starts moving.
To do this, we first calculate the time it takes for the motorcycle to accelerate from rest to a speed of 150 km/h. We can use the equation: final velocity = initial velocity + (acceleration * time).
We know the initial velocity is 0 m/s (rest), the final velocity is 150 km/h (converted to m/s), and the acceleration is given as 6 m/s^2. With these values, we can calculate the time it takes for the motorcycle to reach the speed of 150 km/h.
Then, we calculate the time it takes for the motorcycle to catch up with the car, which is the time it took the car to reach point A plus the time it took the motorcycle to accelerate.
Stage 3: Calculating the distance from point A to the point of overtaking
We can calculate the distance the motorcycle travels from point A to the point of overtaking by multiplying the average speed (150 km/h) by the time it took for the motorcycle to catch up with the car.
By following these steps, you should be able to calculate the distance, s, from point A to the point at which the motorcycle overtakes the car.
120km/h=33.33m/s, 150km/h=41.67m/s
For motorcycle:
v^2=u^2+2*a*s1 (u=0 since at t=2 motorcycle is at rest)
s1=v^2/2a=41.67^2/(2*6)=144.68m
v=u+at1 (u=0)
t1=v/a=41.67/6=6.95s
s2=v(t-6.95) (1)
For car:
s3=v*t (2)
s3=s1+s2
33.33t=144.68+41.67(t-6.95)
t=17.38
and s3=s1+s2 => 579.3=579.3
so the cop overtakes the car 579.3 from A.