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 stopped at the corner (x = 0) starts off in pursuit. the officer accelerates from rest at a = 2.5m/s^2 until reaching a speed of 20 m/s. the officer then slows down at a constant rate until coming alongside the car at x = 360 m, traveling with the same speed as the car
a) how long does it take for the officer to catch up with the motorist?
b)how long does the officer speed up?
c)how far is the officer from the corner and from the car when switching from speeding up to slowing down
d)what is the acceleration of the officer when slowing down
360 /15 = 24 seconds for the whole event (a)
phase 1
cop car Vi = 0, a = 2.5 Vf = 20
v = Vi + a t
20 = 0 + 2.5 t
t = 8 seconds accelerating (b)
x = Xi + Vi t + (1/2) a t^2
x = 0 + 0 t + (1/2)(2.5) (64) = 80 meters accelerating (c)
Phase 2
v = 20 + a t
Vi = 20
Vf = 15
t = 24 - 8 = 16 seconds braking
15 = 20 + a (16)
a = -5 / 15 (d)
what is x and xi?
i can assume x is distance but what is xi if vi is initial velocity
x i means "initial"
in general for constant a
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
That is calculus
if
d^2x/dt^2 = a , a constant
dx/dt = some constant + a t
dx/dt is initial speed Vi when t = 0
so
dx/dt = velocity = v = Vi + a t etc
i means initial in both cases. They are constants of integration assigned by the initial conditions.
typo
a = -5 / 16 (d)
a. Vm * T = 360
15*T = 360
T =
b. V = Vo + a*T = 20
0 + 2.5T = 20
T = 8 s.
c. d = 0.5 a*T^2 = 2.5 * 8^2 = 80 m. from corner.
Vm*T- 80 = 15*8 - 80 = ---m. from car.
To solve this problem, let's break it down into different parts and analyze each step.
a) To find how long it takes for the officer to catch up with the motorist, we need to calculate the time it takes for the officer to cover the distance between them.
We know that the motorist is traveling at a constant speed of 15 m/s. Let's call the time it takes for the officer to catch up with the motorist as t.
For the motorist:
Distance = Speed * Time
Distance = 15 m/s * t
For the officer, we need to calculate the distance covered during acceleration and constant speed. Let's call the distance covered during acceleration as d1 and the distance covered during constant speed as d2.
During acceleration:
d1 = (1/2) * Acceleration * Time^2
20 m/s = (1/2) * 2.5 m/s^2 * t_acceleration^2
During constant speed:
d2 = Speed_after_acceleration * Time_constant_speed
d2 = 20 m/s * t_constant_speed
Total distance covered by the officer:
Total distance = d1 + d2
Total distance = 360 m
Now we can solve the equations:
d1 + d2 = 360 m
Plugging in the value of d1 and d2:
(1/2) * 2.5 m/s^2 * t_acceleration^2 + 20 m/s * t_constant_speed = 360 m
We have two unknowns t_acceleration and t_constant_speed, so we need another equation to solve the system.
b) To find how long the officer speeds up, we can use the equation we obtained above:
20 m/s = (1/2) * 2.5 m/s^2 * t_acceleration^2
Solve for t_acceleration by rearranging the equation:
t_acceleration^2 = (2 * 20 m/s) / 2.5 m/s^2
t_acceleration = sqrt((2 * 20 m/s) / 2.5 m/s^2)
c) To find the distance from the corner and the car when switching from speeding up to slowing down, we need to calculate the distance covered during acceleration. We already know this distance is d1.
d1 = (1/2) * 2.5 m/s^2 * t_acceleration^2
d) To find the acceleration of the officer when slowing down, we can use the equation of motion:
Speed_final^2 = Speed_initial^2 + 2 * Acceleration * Distance
We know the final speed is 15 m/s, initial speed is 20 m/s, and the distance is 360 m. Rearranging the equation, we can solve for acceleration:
Acceleration = (Speed_final^2 - Speed_initial^2) / (2 * Distance)
Plugging in the values:
Acceleration = (15 m/s^2 - 20 m/s^2) / (2 * 360 m)
Now you have the methods to solve each part of the problem. Applying these equations will give you the answers you need.