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) s1= vt - for motorist
s2= (1/2)*a*t^2 - for police officer
when the cop cathes up the motorist:
s1 = s2
So vt = (1/2)*a*t^2
15*t= (1/2)*3*t^2
t = 15*2/3 = 10 sec
b) v = u + a*t
= 0 + 3*10 = 30 m/s
c) distance s1=s2= 15*10 = 150m
corect answer
Why is the acceleration for motorist 0?
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Why time taken is same for motorcycle and police
(a) Well, well, well, looks like we have a high-speed pursuit on our hands! To find out how much time elapses before the police catches up to the motorist, we need to first determine the distance the motorist travels before the police starts the pursuit.
Since the motorist is traveling at a constant speed of 15m/s and the speed limit is 10m/s, we can calculate the distance covered before the police starts as follows:
Distance covered = Speed × Time
Distance covered by motorist = 15m/s × Time
Now, let's look at the police officer. The officer starts from rest and undergoes constant acceleration of 3m/s^2. To catch up to the motorist, the equation we'll need is:
Distance covered = Initial velocity × Time + 0.5 × Acceleration × Time^2
But here's the thing, we want this distance to be equal to the distance covered by the motorist. So we can set up the equation:
Distance covered by police = Distance covered by motorist
(Initial velocity of police × Time + 0.5 × Acceleration of police × Time^2) = (15m/s × Time)
Now, all we need to do is solve for Time. Get your calculators ready!
(b) Once we've determined the time it takes for the police to catch up to the motorist, we can find the officer's speed at that point. To calculate the officer's speed, we'll use the equation:
Officer's speed = Initial velocity + Acceleration × Time
Plug in the values, do the math, and voila!
(c) Finally, to calculate the total distance each vehicle has traveled at that point, we need to add up the distance covered by each of them. For the motorist, it will be:
Distance covered by motorist = Speed of motorist × Time
And for the police officer, it will be:
Distance covered by police = Initial velocity × Time + 0.5 × Acceleration × Time^2
Once again, grab your calculators and let's crunch those numbers!
Remember, in the world of high-speed pursuits, calculations can be intense but comedy can lighten the mood. Stay safe out there, folks!
To solve this problem, we need to use the equations of motion and apply them to both the motorist and the police officer. Let's break down each part of the problem and solve it step by step:
(a) How much time elapses before the police officer catches up with the motorist?
We know that the motorist's speed is constant at 15 m/s. Let's assume that the motorist has traveled a distance "d" when the police officer starts pursuit.
Using the equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
For the motorist:
15t = d
For the police officer:
(1/2) * 3 * t^2 = d
Setting the two equations equal to each other, we can solve for "t":
15t = (1/2) * 3 * t^2
Simplifying the equation:
15t = (3/2) * t^2
Rearranging to isolate "t":
(3/2) * t^2 - 15t = 0
Factoring out "t":
t * (3/2 * t - 15) = 0
From this equation, we have two solutions: t = 0 (which is not physically meaningful in this context) or (3/2 * t - 15) = 0
Solving for "t":
(3/2 * t - 15) = 0
(3/2 * t) = 15
t = 10 seconds
So, it takes the police officer 10 seconds to catch up with the motorist.
(b) What is the officer's speed at that point?
To find the officer's speed at the time of catching up, we need to use the equation:
final velocity = initial velocity + acceleration * time
The initial velocity of the police officer is 0 m/s (since he started from rest), and acceleration is 3 m/s^2 (given in the problem).
Plugging in the values:
final velocity = 0 + 3 * 10
final velocity = 30 m/s
So, the officer's speed at that point is 30 m/s.
(c) What is the total distance each vehicle has traveled at that point?
To find the total distance traveled by each vehicle, we need to find the distance covered by each vehicle until the time of catching up.
For the motorist, we know that the speed is 15 m/s, and the time taken is 10 seconds:
distance = speed * time
distance = 15 * 10
distance = 150 meters
So, the motorist traveled 150 meters before being caught.
For the police officer, we need to calculate the distance covered using the equation of motion:
distance = (initial velocity * time) + (1/2) * acceleration * time^2
Taking the time as 10 seconds:
distance = (0 * 10) + (1/2) * 3 * (10^2)
distance = (1/2) * 3 * 100
distance = 150 meters
So, the police officer also traveled 150 meters before catching up.
In conclusion:
(a) The time elapsed before the police officer catches up is 10 seconds.
(b) The officer's speed at that point is 30 m/s.
(c) The total distance traveled by each vehicle at that point is 150 meters.