Question

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

Answer 1

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)

Answer 2

what is x and xi?
i can assume x is distance but what is xi if vi is initial velocity

Answer 3

x i means "initial"

Answer 4

in general for constant a
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2

Answer 5

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

Answer 6

i means initial in both cases. They are constants of integration assigned by the initial conditions.

Answer 7

typo
a = -5 / 16 (d)

Answer 8

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.