A motorist traveling with a constant speed of 15m/s passes a school-crossing corner, where the speed limit is 10m/s. Just as the motorist passes, a police officer on a motorcycle at the corner starts off in pursuit with constant acceleration of 3m/s^2. (a)How much time elapses before the police catches up the motorist? (b)What is the officer's speed limit at that point? (c)What is the total distance each vehicle has traveled at that point?
a. they both go the same distance.
distancecar=15*t
distanccop=1/2 * 3*t^2 set them equal, solve for t.
officersspeed= 3*t
distance= 15t
1. A motorist travelling 45 m/s [ abaout 75 mi/h] passes a stationary motorcycle police officer. 2,5 second after motorist passes, the police officer sarts to move and accelerates in pursuit of the speeding motorist. The motorcycle has constant acceleration of 3,6 m/s2.
a) How fast will the police officer be travelling when he overtakes the car ?draw curves of x versus t for both the motorcycle and the car, taking t-0 at the moment the car passes the stationary police officer.
b) Suppose that for reason of safety the policeman does not exceed a maximum speed of 45 m/s [about 100mi/h]. how long will it then take him to overtake the car, and how far will he have traveled ?
To solve this problem, we can use the kinematic equations of motion. Let's call the time at which the police catches up to the motorist as T. We will also assume that the initial position of both the motorist and the police officer are the same.
(a) To find the time taken by the police to catch up to the motorist, we can use the equation:
Distance = Speed × Time + (1/2) × Acceleration × Time^2
For the motorist:
Distance = 15 × T
For the police officer:
Distance = 10 × T + (1/2) × 3 × T^2
Since the distances are the same when they catch up, we can equate the two equations:
15 × T = 10 × T + (1/2) × 3 × T^2
Simplifying the equation:
15T = 10T + (3/2)T^2
(3/2)T^2 - 5T = 0
Factoring out T:
T((3/2)T - 5) = 0
So, T = 0 or T = 10/3 seconds.
Since the time cannot be zero for the police to catch up, the time taken for the police to catch up to the motorist is T = 10/3 seconds.
(b) To find the officer's speed at the point of catching up, we can use the equation:
Speed = Initial Speed + (Acceleration × Time)
For the police officer:
Speed = 10 + (3 × (10/3))
Speed = 20 m/s
Therefore, the officer's speed at the point of catching up is 20 m/s.
(c) To find the total distance traveled by each vehicle at that point, we can use the equation:
Distance = Initial Distance + (Speed × Time) + (1/2) × Acceleration × Time^2
For the motorist:
Distance = 15 × (10/3)
For the police officer:
Distance = 10 × (10/3) + (1/2) × 3 × (10/3)^2
Simplifying the equations:
Distance for the motorist = 150/3 = 50 m
Distance for the police officer = 100/3 + (1/2) × 3 × (100/9) = 100/3 + 100/3 = 200/3 = 66.67 m
Therefore, at the point of catching up, the motorist has traveled 50 meters, and the police officer has traveled 66.67 meters.
To solve this problem, we need to analyze the motion of both the motorist and the police officer separately, and then find the point where they intersect.
(a) How much time elapses before the police catches up to the motorist?
Let's denote the time elapsed as 't'. For the motorist, the distance covered is given by the formula: distance = speed x time.
So, the distance covered by the motorist is: distance_m = 15t.
For the police officer, we need to consider that the initial speed is 0 when the motorist passes the corner and the acceleration is 3 m/s^2. The distance traveled by the police officer can be calculated using the formula: distance = initial velocity x time + (1/2) x acceleration x time^2.
Thus, the distance covered by the police officer is: distance_p = (1/2) x 3t^2.
To find the time when they intersect, we set the distances equal to each other: 15t = (1/2) x 3t^2.
Simplifying the equation, we get: 15t = (3/2)t^2.
Rearranging the equation, we have: (3/2)t^2 - 15t = 0.
Factoring out a 't', we get: t((3/2)t - 15) = 0.
Setting each factor equal to zero, we find two possibilities: t = 0 (which we ignore since we are finding the time when they meet) and (3/2)t - 15 = 0.
Solving for t in the second equation, we find: (3/2)t = 15. Dividing by (3/2), we get t = 10 seconds.
Hence, the police officer catches up to the motorist after 10 seconds.
(b) What is the officer's speed at that point?
To find the officer's speed at that point, we need to calculate the velocity of the police officer using a constant acceleration formula. The formula for velocity is: velocity = initial velocity + acceleration x time.
Given that the initial velocity of the police officer is 0 and the acceleration is 3 m/s^2, we can substitute the values into the formula to find: velocity = 0 + 3 x 10.
Thus, the officer's speed is 30 m/s.
Therefore, the officer's speed limit at that point is 30 m/s.
(c) What is the total distance each vehicle has traveled at that point?
To calculate the total distance, we add up the distance covered by each vehicle.
For the motorist, we know the distance covered is: distance_m = 15t, where t = 10 seconds. Substituting the values, we find: distance_m = 15 x 10 = 150 meters.
For the police officer, we have the distance covered as: distance_p = (1/2) x 3t^2, where t = 10 seconds. Substituting the values, we find: distance_p = (1/2) x 3 x 10^2 = 150 meters.
Therefore, at the point when the police officer catches up to the motorist, both vehicles have traveled a total distance of 150 meters.